Class 10 Optional Mathematics Polynomial
Exercise 2.1 Remainder Theorem, Exercise 2.2 Factor Theorem and Exercise 2.3 Factorization
45 solved questions with English and Nepali questions, step by step solutions and frequently asked questions for the Nepal Grade 10 SEE.
1. Class 10 Optional Mathematics Polynomial: Unit Overview and Key Formulas
Unit 2: Polynomial (Optional Mathematics, Grade 10, SEE, Nepal)
Class 10 Optional Mathematics Polynomial is the second unit of the Grade 10 Optional Mathematics course prescribed by the Curriculum Development Centre (CDC), Nepal. These class 10 optional math polynomial notes solve every question of Exercise 2.1, Exercise 2.2 and Exercise 2.3.
Every question is written as in the textbook: the main question first (Q.1, Q.2, Q.3 …) with its parts (a, b, c …) listed under it and the Nepali question in brackets. Multiple choice solutions, definitions and proofs also have Nepali in brackets, and the other solutions are in English because the steps and symbols are the same in both languages.
- Exercise 2.1 Remainder Theorem: division algorithm, finding remainders, unknown constants and equal remainders.
- Exercise 2.2 Factor Theorem: proving factors, finding constants and the number to be added or subtracted.
- Exercise 2.3 Factorization and Solving Polynomial Equations: zeros, forming polynomials, synthetic division, factorizing cubics and solving cubic equations.
Key formulas to remember
For more chapters, visit the complete SEE Class 10 Optional Mathematics notes. You can also read about the polynomial remainder theorem and synthetic division for extra understanding.
2. Exercise 2.1 Remainder Theorem: Class 10 Optional Mathematics Polynomial
Exercise 2.1 Remainder Theorem: Multiple Choice Questions (बहुवैकल्पिक प्रश्न)
Q.1 Tick (✓) the correct option for the given questions:
(दिइएका प्रश्नहरूको उपयुक्त विकल्पमा ठिक चिह्न (✓) लगाउनुहोस् : )
A. If the polynomials \(f(x)=x^3+x^2+x+1\) and \(g(x)=x+1\), what will be the degree of \(f(x)\div g(x)\)? (बहुपदीय \(f(x)=x^3+x^2+x+1\) र \(g(x)=x+1\) भए \(f(x)\div g(x)\) को डिग्री कति होला ?)
- a) 1
- b) 2
- c) 3
- d) 4
Correct Answer: b) 2
Solution:
- When polynomials are divided, the degree of the quotient equals the degree of the dividend minus the degree of the divisor.
(भाग गर्दा भागफलको डिग्री = \(f(x)\) को डिग्री \(-\) \(g(x)\) को डिग्री = \(3-1=2\) हुन्छ ।)
B. When the polynomial \(f(x)\) is divided by the divisor \(d(x)\), the quotient is \(q(x)\) and the remainder is \(R\). Which of the following relations between \(f(x)\), \(d(x)\), \(q(x)\) and \(R\) is correct? (बहुपदीय \(f(x)\) लाई भाजक \(d(x)\) ले भाग गर्दा भागफल \(q(x)\) र शेष \(R\) हुन्छ भने \(f(x)\), \(d(x)\), \(q(x)\) र \(R\) को सम्बन्ध तलका मध्ये कुन सही हुन्छ ?)
- a) \(f(x)=d(x)\times q(x)+R\)
- b) \(f(x)=q(x)\times q(x)+R\)
- c) \(f(x)=R\times q(x)+d(x)\)
- d) \(f(x)=d(x)\times R+R\)
Correct Answer: a) \(f(x)=d(x)\times q(x)+R\)
Solution:
- By the polynomial division algorithm, Dividend \(=\) (Divisor \(\times\) Quotient) \(+\) Remainder.
(भाग सम्बन्धी नियम अनुसार भाज्य \(=\) (भाजक \(\times\) भागफल) \(+\) शेष हुन्छ ।)
C. What will be the remainder when the polynomial \(f(x)\) is divided by \(x-a\)? (बहुपदीय \(f(x)\) लाई \(x-a\) ले भाग गर्दा शेष कति होला ?)
- a) 0
- b) 1
- c) \(f(x)\)
- d) \(f(a)\)
Correct Answer: d) \(f(a)\)
Solution:
- By the Remainder Theorem, when \(f(x)\) is divided by \((x-a)\), the remainder is \(f(a)\).
(शेष साध्य अनुसार \(f(x)\) लाई \((x-a)\) ले भाग गर्दा आउने शेष \(f(a)\) हुन्छ ।)
D. What will be the remainder when the polynomial \(f(x)\) is divided by \(ax+b\)? (बहुपदीय \(f(x)\) लाई \(ax+b\) ले भाग गर्दा शेष कति होला ?)
- a) \(f\left(\frac{a}{b}\right)\)
- b) \(f\left(-\frac{a}{b}\right)\)
- c) \(f\left(-\frac{b}{a}\right)\)
- d) \(f\left(\frac{b}{a}\right)\)
Correct Answer: c) \(f\left(-\frac{b}{a}\right)\)
Solution:
- Set the divisor equal to zero.
- By the Remainder Theorem, the remainder is \(R=f\left(-\frac{b}{a}\right)\).
(भाजक \(ax+b=0\) मान्दा \(x=-\frac{b}{a}\) हुन्छ । तसर्थ शेष साध्य अनुसार शेष \(R=f\left(-\frac{b}{a}\right)\) हुन्छ ।)
E. What will be the remainder when the polynomial \((x^3-x^2+1)\) is divided by \((x-1)\)? (बहुपदीय \((x^3-x^2+1)\) लाई \((x-1)\) ले भाग गर्दा शेष कति हुन्छ ?)
- a) 0
- b) 1
- c) 2
- d) 3
Correct Answer: b) 1
Solution:
- Let \(f(x)=x^3-x^2+1\). Comparing \((x-1)\) with \((x-a)\) gives \(a=1\).
(\(f(x)=x^3-x^2+1\) मा \(x=1\) राख्दा शेष \(R=1\) आउँछ ।)
F. If the remainder is \(4\) when the polynomial \(x^3+3x^2-kx+4\) is divided by \(x-2\), what is the value of \(k\)? (बहुपदीय \(x^3+3x^2-kx+4\) लाई \(x-2\) ले भाग गर्दा शेष \(4\) रहन्छ भने \(k\) को मान कति होला ?)
- a) 10
- b) 8
- c) 4
- d) 2
Correct Answer: a) 10
Solution:
- Let \(f(x)=x^3+3x^2-kx+4\). The divisor is \(x-2\), so \(a=2\), and \(R=4\).
(शेष साध्य अनुसार \(f(2)=4\) राखी हल गर्दा \(k=10\) प्राप्त हुन्छ ।)
Exercise 2.1 Remainder Theorem: Written Questions (लेख्ने प्रश्न)
Statement of the Remainder Theorem
If a polynomial \(f(x)\) of degree \(n\ge 1\) is divided by a linear divisor \((x-a)\), then the remainder is equal to \(f(a)\).
(यदि डिग्री \(n\ge 1\) को बहुपदीय \(f(x)\) लाई रेखीय भाजक \((x-a)\) ले भाग गर्दा आउने शेष \(f(a)\) हुन्छ ।)
Proof
Let \(q(x)\) be the quotient and \(R\) the remainder when \(f(x)\) is divided by \((x-a)\). The degree of the remainder is less than the degree of \(x-a\), so \(R\) is a constant.
By the polynomial division algorithm:
\[f(x)=(x-a)\cdot q(x)+R\]This relation holds for all values of \(x\). Substitute \(x=a\):
\[f(a)=(a-a)\cdot q(a)+R=0\cdot q(a)+R=R\]Hence the remainder is \(R=f(a)\).
(मानौँ \(f(x)\) लाई \((x-a)\) ले भाग गर्दा भागफल \(q(x)\) र शेष \(R\) (स्थिर) हुन्छ । भाग सम्बन्धी नियम अनुसार \(f(x)=(x-a)\cdot q(x)+R\) हुन्छ, जुन \(x\) का सबै मानका लागि सत्य हुन्छ । यसमा \(x=a\) राख्दा \(f(a)=0\cdot q(a)+R=R\) आउँछ । अतः शेष \(R=f(a)\) हुन्छ ।)
Q.3 Using the remainder theorem in the given expressions, find the remainder:
- a. \((x^3-x^2+1)\div(x-2)\)
- b. \((x^3+5x^2-6x-16)\div(x+1)\)
- c. \((6x^3+4x^2+3x+4)\div(3x-4)\)
- d. \((8x^4+5x^3-2x^2+7x-1)\div(2x+6)\)
(दिइएको अवस्थामा शेष साध्य प्रयोग गरी शेष पत्ता लगाउनुहोस् : )
Step 1: Identify
Dividend \(f(x)=x^3-x^2+1\) and divisor \(x-2\). Comparing with \(x-a\) gives \(a=2\).
Step 2: Apply the Remainder Theorem
\(R=f(2)\)
Step 3: Substitute and simplify
\[f(2)=(2)^3-(2)^2+1=8-4+1=5\]Step 1: Identify
\(f(x)=x^3+5x^2-6x-16\) and divisor \(x+1\). Comparing with \(x-a\) gives \(a=-1\).
Step 2: Apply the Remainder Theorem
\(R=f(-1)\)
Step 3: Substitute and simplify
\[\begin{aligned}f(-1) &= (-1)^3+5(-1)^2-6(-1)-16 \\ &= -1+5+6-16 \\ &= -6\end{aligned}\]Step 1: Identify
\(f(x)=6x^3+4x^2+3x+4\) and divisor \(3x-4\).
Step 2: Find the value of \(x\) for the divisor
\[\begin{aligned}& 3x-4=0 \\ &\implies 3x=4 \\ &\implies x=\frac{4}{3}\end{aligned}\]By the Remainder Theorem, \(R=f\left(\frac{4}{3}\right)\).
Step 3: Substitute and simplify
\[\begin{aligned}f\left(\frac{4}{3}\right) &= 6\left(\frac{4}{3}\right)^3+4\left(\frac{4}{3}\right)^2+3\left(\frac{4}{3}\right)+4 \\ &= 6\cdot\frac{64}{27}+4\cdot\frac{16}{9}+4+4 \\ &= \frac{128}{9}+\frac{64}{9}+8 \\ &= \frac{192}{9}+8 \\ &= \frac{64}{3}+8 \\ &= \frac{88}{3}\end{aligned}\]Step 1: Identify
\(f(x)=8x^4+5x^3-2x^2+7x-1\) and divisor \(2x+6\).
Step 2: Find the value of \(x\)
\[2x+6=0 \implies 2x=-6 \implies x=-3\]By the Remainder Theorem, \(R=f(-3)\).
Step 3: Substitute and simplify
\[\begin{aligned}f(-3) &= 8(-3)^4+5(-3)^3-2(-3)^2+7(-3)-1 \\ &= 8(81)+5(-27)-2(9)-21-1 \\ &= 648-135-18-21-1 \\ &= 473\end{aligned}\]Step 1: Identify
\(x+2=0\) gives \(a=-2\), and the remainder is \(R=16\).
Step 2: Apply the Remainder Theorem
\(f(-2)=16\)
Step 3: Substitute \(x=-2\) and solve for \(k\)
\[2(-2)^3+3(-2)^2-k(-2)+4=16\]\[\begin{aligned}& 2(-8)+3(4)+2k+4=16 \\ &\implies -16+12+2k+4=16 \\ &\implies 2k=16 \\ &\implies k=8\end{aligned}\]Step 1: Identify
\[2x-1=0 \implies x=\frac{1}{2}\]The remainder is \(R=-2\).
Step 2: Apply the Remainder Theorem
\(f\left(\frac{1}{2}\right)=-2\)
Step 3: Substitute \(x=\frac{1}{2}\) and solve for \(k\)
\[8\left(\frac12\right)^3+4\left(\frac12\right)^2+(k-3)\left(\frac12\right)-7=-2\]\[\begin{aligned}& 1+1+\frac{k-3}{2}-7=-2 \\ &\implies -5+\frac{k-3}{2}=-2 \\ &\implies \frac{k-3}{2}=3 \\ &\implies k-3=6 \\ &\implies k=9\end{aligned}\]Step 1: Identify
\(x+3=0\) gives \(a=-3\), and the remainder is \(R=10\), so \(f(-3)=10\).
Step 2: Substitute \(x=-3\) and solve for \(p\)
\[(-3)^3-(p-2)(-3)^2-p(-3)+28=10\]\[\begin{aligned}& -27-(p-2)(9)+3p+28=10 \\ &\implies -27-9p+18+3p+28=10 \\ &\implies -6p+19=10 \\ &\implies -6p=-9 \\ &\implies p=\frac{3}{2}\end{aligned}\]Step 1: Identify
Divisor \(x+2\) gives \(a=-2\). The remainders are equal, so \(f(-2)=g(-2)\).
Step 2: Calculate \(f(-2)\)
\[\begin{aligned}f(-2) &= 2(-2)^3-(k-2)(-2)^2-7(-2)+14 \\ &= 2(-8)-4(k-2)+14+14 \\ &= -16-4k+8+28 \\ &= 20-4k\end{aligned}\]Step 3: Calculate \(g(-2)\)
\[\begin{aligned}g(-2) &= 3(-2)^3+8(-2)^2-k(-2)+6 \\ &= 3(-8)+8(4)+2k+6 \\ &= -24+32+2k+6 \\ &= 14+2k\end{aligned}\]Step 4: Equate and solve
\[20-4k=14+2k \implies 6=6k \implies k=1\]Step 1: Identify
\[2x-1=0 \implies x=\frac{1}{2}\]The remainders are equal, so \(f\left(\frac12\right)=g\left(\frac12\right)\).
Step 2: Calculate \(f\left(\frac12\right)\)
\[\begin{aligned}f\left(\frac12\right) &= 2\left(\frac18\right)+h\left(\frac14\right)-5+8 \\ &= \frac14+\frac{h}{4}+3 \\ &= \frac{h+13}{4}\end{aligned}\]Step 3: Calculate \(g\left(\frac12\right)\)
\[\begin{aligned}g\left(\frac12\right) &= 6\left(\frac18\right)-h\left(\frac14\right)+6 \\ &= \frac34-\frac{h}{4}+6 \\ &= \frac{27-h}{4}\end{aligned}\]Step 4: Equate and solve
\[\begin{aligned}& \frac{h+13}{4}=\frac{27-h}{4} \\ &\implies h+13=27-h \\ &\implies 2h=14 \\ &\implies h=7\end{aligned}\]Step 1: Identify
- Divided by \((x+2)\), the remainder is \(24\), so \(f(-2)=24\).
- Divided by \((x-3)\), the remainder is \(54\), so \(f(3)=54\).
Step 2: Form equation (i) from \(f(-2)=24\)
\[\begin{aligned}& (-2)^3+m(-2)^2-n(-2)+6=24 \\ &\implies -8+4m+2n+6=24 \\ &\implies 4m+2n=26\end{aligned}\]Divide by \(2\):
\[2m+n=13 \qquad \text{(i)}\]Step 3: Form equation (ii) from \(f(3)=54\)
\[\begin{aligned}& (3)^3+m(3)^2-n(3)+6=54 \\ &\implies 27+9m-3n+6=54 \\ &\implies 9m-3n=21\end{aligned}\]Divide by \(3\):
\[3m-n=7 \qquad \text{(ii)}\]Step 4: Solve (i) and (ii)
Add (i) and (ii):
\[\begin{aligned}& (2m+n)+(3m-n)=13+7 \\ &\implies 5m=20 \\ &\implies m=4\end{aligned}\]Substitute \(m=4\) in (i):
\[2(4)+n=13 \implies n=5\]3. Exercise 2.2 Factor Theorem: Class 10 Optional Mathematics Polynomial
Exercise 2.2 Factor Theorem: Multiple Choice Questions (बहुवैकल्पिक प्रश्न)
Q.1 Tick (✓) the correct option from the given questions:
(दिइएका प्रश्नहरूको उपयुक्त विकल्पमा ठिक चिह्न (✓) लगाउनुहोस् : )
A. When a polynomial \(f(x)\) is divided by a factor \((x-a)\), what is the value of \(f(a)\)? (बहुपदीय \(f(x)\) को एउटा गुणनखण्ड \((x-a)\) ले \(f(x)\) लाई भाग गर्दा \(f(a)\) को मान कति हुन्छ ?)
- a) 0
- b) 1
- c) \(a\)
- d) \(-a\)
Correct Answer: a) 0
Solution:
- By the Factor Theorem, if \((x-a)\) is a factor of \(f(x)\), the remainder is \(f(a)=0\).
(गुणनखण्ड साध्य अनुसार यदि \((x-a)\) बहुपदीय \(f(x)\) को गुणनखण्ड हो भने \(f(a)=0\) हुन्छ ।)
B. When the polynomial \(f(x)\) is divided by a linear factor, the remainder is \(f(-k)\). What is that factor? (बहुपदीय \(f(x)\) लाई एउटा रेखीय गुणनखण्डले भाग गर्दा शेष \(f(-k)\) भए उक्त गुणनखण्ड कति हुन्छ ?)
- a) \(x-a\)
- b) \(x+k\)
- c) \(x-k\)
- d) \(x+a\)
Correct Answer: b) \(x+k\)
Solution:
- Since the remainder is \(f(-k)\), the divisor must vanish at \(x=-k\).
- Thus the factor is \((x+k)\).
(शेष \(f(-k)\) प्राप्त हुनका लागि गुणनखण्ड \(x+k\) हुनुपर्छ, जहाँ \(x+k=0\implies x=-k\) हुन्छ ।)
C. In which condition does the polynomial \(f(x)\) have a factor \((cx-d)\)? (कुन अवस्थामा बहुपदीय \(f(x)\) को एउटा गुणनखण्ड \((cx-d)\) हुन्छ ?)
- a) \(f\left(\frac{c}{d}\right)=0\)
- b) \(f\left(-\frac{c}{d}\right)=0\)
- c) \(f\left(\frac{d}{c}\right)=0\)
- d) \(f\left(-\frac{d}{c}\right)=0\)
Correct Answer: c) \(f\left(\frac{d}{c}\right)=0\)
Solution:
- Set the factor equal to zero.
- By the Factor Theorem, \(f\left(\frac{d}{c}\right)=0\).
(गुणनखण्ड \(cx-d=0\implies x=\frac{d}{c}\) मान्दा \(f\left(\frac{d}{c}\right)=0\) हुनुपर्छ ।)
D. If \((x-1)\) is a factor of the polynomial \(x^3+x^2+p+1\), what is the value of \(p\)? (यदि बहुपदीय \(x^3+x^2+p+1\) को एउटा गुणनखण्ड \((x-1)\) भए \(p\) को मान कति हुन्छ ?)
- a) \(-5\)
- b) \(-4\)
- c) \(-3\)
- d) \(-1\)
Correct Answer: c) \(-3\)
Solution:
- Let \(f(x)=x^3+x^2+p+1\). Since \((x-1)\) is a factor, \(f(1)=0\).
(\(f(1)=0\) राखेर हल गर्दा \(p=-3\) आउँछ ।)
E. If \((x+3)\) is a factor of the polynomial \(3x^3+mx^2-7x-3\), what is the value of \(m\)? (यदि बहुपदीय \(3x^3+mx^2-7x-3\) को एउटा गुणनखण्ड \((x+3)\) भए \(m\) को मान कति हुन्छ ?)
- a) 2
- b) 3
- c) 5
- d) 7
Correct Answer: d) 7
Solution:
- Let \(f(x)=3x^3+mx^2-7x-3\). Since \((x+3)\) is a factor, \(f(-3)=0\).
(\(f(-3)=0\) राखेर हल गर्दा \(m=7\) प्राप्त हुन्छ ।)
Exercise 2.2 Factor Theorem: Written Questions (लेख्ने प्रश्न)
Statement of the Factor Theorem
Let \(f(x)\) be a polynomial of degree \(n\ge 1\). A linear polynomial \((x-a)\) is a factor of \(f(x)\) if and only if \(f(a)=0\).
(मानौँ \(f(x)\) डिग्री \(n\ge 1\) को बहुपदीय हो । \((x-a)\) बहुपदीय \(f(x)\) को गुणनखण्ड हुन्छ यदि र मात्र यदि \(f(a)=0\) हुन्छ ।)
Proof
By the Remainder Theorem, dividing \(f(x)\) by \((x-a)\) gives
\[f(x)=(x-a)\cdot q(x)+f(a)\]where \(q(x)\) is the quotient and \(f(a)\) is the remainder.
- Case 1: Assume \(f(a)=0\). Then \(f(x)=(x-a)\cdot q(x)\), so \((x-a)\) divides \(f(x)\) exactly. Hence \((x-a)\) is a factor of \(f(x)\).
- Case 2: Conversely, assume \((x-a)\) is a factor. Then \(f(x)=(x-a)\cdot q(x)\) for some \(q(x)\). Substituting \(x=a\):
((१) \(f(a)=0\) मान्दा \(f(x)=(x-a)\cdot q(x)\) हुन्छ, अर्थात् \((x-a)\) ले \(f(x)\) लाई निश्शेष भाग गर्छ, त्यसैले \((x-a)\) गुणनखण्ड हो । (२) विपरीत रूपमा \((x-a)\) गुणनखण्ड हो भने \(f(x)=(x-a)\cdot q(x)\) हुन्छ र \(x=a\) राख्दा \(f(a)=0\) आउँछ ।)
Proof
Let \(f(x)\) be a polynomial of degree \(n\ge 1\). Since \((x-a)\) is a factor, \(f(x)\) can be written as the product of \((x-a)\) and a quotient polynomial \(q(x)\):
\[f(x)=(x-a)\cdot q(x)\]Substitute \(x=a\):
\[f(a)=(a-a)\cdot q(a)=0\cdot q(a)=0\]By the Remainder Theorem the remainder is \(R=f(a)\). Since \(f(a)=0\), the remainder is \(0\).
(\((x-a)\) गुणनखण्ड भएकाले \(f(x)=(x-a)\cdot q(x)\) हुन्छ । यसमा \(x=a\) राख्दा \(f(a)=0\cdot q(a)=0\) आउँछ । शेष साध्य अनुसार शेष \(R=f(a)=0\) हुन्छ ।)
Q.4 Using the Factor Theorem, prove the following in the given cases:
- a. \(x-3\) is a factor of the polynomial \(f(x)=2x^2-11x+15\)
- b. \(2x-1\) is a factor of the polynomial \(f(x)=2x^3-7x^2+13x-5\)
- c. \(2x+1\) is a factor of the polynomial \(f(x)=2x^3-5x^2+7x+5\)
(दिइएको अवस्थामा गुणनखण्ड साध्य प्रयोग गरी प्रमाणित गर्नुहोस् : )
Step 1: Identify
Proposed factor \(x-3\) gives \(a=3\).
Step 2: Principle
By the Factor Theorem, \((x-3)\) is a factor of \(f(x)\) if \(f(3)=0\).
Step 3: Evaluate \(f(3)\)
\[\begin{aligned}f(3) &= 2(3)^2-11(3)+15 \\ &= 2(9)-33+15 \\ &= 18-33+15 \\ &= 0\end{aligned}\]Since \(f(3)=0\), \((x-3)\) is a factor of \(f(x)\).
Step 1: Identify
\[2x-1=0 \implies x=\frac{1}{2}\]Step 2: Principle
By the Factor Theorem, \((2x-1)\) is a factor of \(f(x)\) if \(f\left(\frac12\right)=0\).
Step 3: Evaluate \(f\left(\frac12\right)\)
\[\begin{aligned}f\left(\frac12\right) &= 2\left(\frac12\right)^3-7\left(\frac12\right)^2+13\left(\frac12\right)-5 \\ &= \frac14-\frac74+\frac{13}{2}-5 \\ &= \frac{1-7+26-20}{4} \\ &= \frac{0}{4} \\ &= 0\end{aligned}\]Since \(f\left(\frac12\right)=0\), \((2x-1)\) is a factor of \(f(x)\).
Step 1: Identify
\[2x+1=0 \implies x=-\frac{1}{2}\]Step 2: Principle
By the Factor Theorem, \((2x+1)\) is a factor of \(f(x)\) if \(f\left(-\frac12\right)=0\).
Step 3: Evaluate \(f\left(-\frac12\right)\)
\[\begin{aligned}f\left(-\frac12\right) &= 2\left(-\frac18\right)-5\left(\frac14\right)-\frac72+5 \\ &= -\frac14-\frac54-\frac72+5 \\ &= \frac{-1-5-14+20}{4} \\ &= \frac{0}{4} \\ &= 0\end{aligned}\]Since \(f\left(-\frac12\right)=0\), \((2x+1)\) is a factor of \(f(x)\).
Step 1: Apply the Factor Theorem
\(x+2\) gives \(a=-2\). Since \((x+2)\) is a factor, \(f(-2)=0\).
Step 2: Substitute \(x=-2\)
\[4(-2)^2+k(-2)+8=0\]\[\begin{aligned}& 4(4)-2k+8=0 \\ &\implies 24-2k=0 \\ &\implies 2k=24 \\ &\implies k=12\end{aligned}\]Step 1: Apply the Factor Theorem
\(x+1\) gives \(a=-1\). Since \((x+1)\) is a factor, \(f(-1)=0\).
Step 2: Substitute \(x=-1\)
\[2(-1)^3-k(-1)^2-8(-1)+5=0\]\[\begin{aligned}& 2(-1)-k+8+5=0 \\ &\implies 11-k=0 \\ &\implies k=11\end{aligned}\]Step 1: Form the new polynomial
Let \(k\) be the number to be added. Then \(g(x)=f(x)+k=x^3+6x^2+11x+1+k\).
Step 2: Apply the Factor Theorem
\(x+1\) gives \(a=-1\), so \(g(-1)=0\).
\[(-1)^3+6(-1)^2+11(-1)+1+k=0\]\[\begin{aligned}& -1+6-11+1+k=0 \\ &\implies -5+k=0 \\ &\implies k=5\end{aligned}\]Step 1: Form the new polynomial
Let \(k\) be the number to be added. Then \(g(x)=f(x)+k=x^3+x^2-17x+7+k\).
Step 2: Apply the Factor Theorem
\(x-3\) gives \(a=3\), so \(g(3)=0\).
\[(3)^3+(3)^2-17(3)+7+k=0\]\[\begin{aligned}& 27+9-51+7+k=0 \\ &\implies -8+k=0 \\ &\implies k=8\end{aligned}\]Step 1: Form the new polynomial
Let \(k\) be the number to be subtracted. Then \(g(x)=f(x)-k=x^3+9x^2+26x+30-k\).
Step 2: Apply the Factor Theorem
\(x+4\) gives \(a=-4\), so \(g(-4)=0\).
\[(-4)^3+9(-4)^2+26(-4)+30-k=0\]\[\begin{aligned}& -64+144-104+30-k=0 \\ &\implies 6-k=0 \\ &\implies k=6\end{aligned}\]4. Exercise 2.3 Factorization and Solving Polynomial Equations: Class 10 Optional Mathematics Polynomial
Exercise 2.3 Factorization and Solving Polynomial Equations: Written Questions (लेख्ने प्रश्न)
Step 1: Test each value by evaluating \(f(x)\)
Test \(x=1\):
\[f(1)=1-6+11-6=0\]So \(x=1\) is a root and \((x-1)\) is a factor.
Test \(x=-1\):
\[\begin{aligned}f(-1) &= (-1)^3-6(-1)^2+11(-1)-6 \\ &= -1-6-11-6 \\ &= -24\neq 0\end{aligned}\]So \(x=-1\) is not a root.
Test \(x=-2\):
\[\begin{aligned}f(-2) &= (-2)^3-6(-2)^2+11(-2)-6 \\ &= -8-24-22-6 \\ &= -60\neq 0\end{aligned}\]So \(x=-2\) is not a root.
Test \(x=2\):
\[f(2)=8-24+22-6=0\]So \(x=2\) is a root and \((x-2)\) is a factor.
Test \(x=-3\):
\[\begin{aligned}f(-3) &= (-3)^3-6(-3)^2+11(-3)-6 \\ &= -27-54-33-6 \\ &= -120\neq 0\end{aligned}\]So \(x=-3\) is not a root.
Test \(x=3\):
\[f(3)=27-54+33-6=0\]So \(x=3\) is a root and \((x-3)\) is a factor.
Step 2: State the factors
The zeros are \(1,\,2,\,3\) and the factors are \((x-1),\,(x-2),\,(x-3)\).
Step 1: Write the factors
If \(a\) is a zero of \(f(x)\), then \((x-a)\) is a factor. The factors are \((x-2),\,(x-4),\,(x-6)\).
Step 2: Multiply the factors
\[f(x)=(x-2)(x-4)(x-6)\]First multiply two factors:
\[(x-2)(x-4)=x^2-4x-2x+8=x^2-6x+8\]Now multiply by \((x-6)\):
\[\begin{aligned}f(x) &= (x^2-6x+8)(x-6) \\ &= x(x^2-6x+8)-6(x^2-6x+8) \\ &= x^3-6x^2+8x-6x^2+36x-48 \\ &= x^3-12x^2+44x-48\end{aligned}\]Q.3 Factorize:
- a. \(x^3+6x^2+11x+6\)
- b. \(x^3-3x^2-4x+12\)
- c. \(x^3-8x^2+19x-12\)
- d. \((x+2)(x^2-11x+48)-120\)
- e. \((x-3)(3x^2-x^2-10x+2)-10\)
(खण्डीकरण गर्नुहोस् : )
Step 1: Find a zero
The constant term is \(6\), so try \(\pm1,\pm2,\pm3,\pm6\). Let \(f(x)=x^3+6x^2+11x+6\).
\[\begin{aligned}f(-1) &= (-1)^3+6(-1)^2+11(-1)+6 \\ &= -1+6-11+6 \\ &= 0\end{aligned}\]So \((x+1)\) is a factor.
Step 2: Synthetic division by \(-1\)
\[\begin{array}{r|rrrr} -1 & 1 & 6 & 11 & 6 \\ & & -1 & -5 & -6 \\ \hline & 1 & 5 & 6 & 0 \end{array}\]The quotient is \(q(x)=x^2+5x+6\).
Step 3: Factorize the quotient
\[\begin{aligned}x^2+5x+6 &= x^2+3x+2x+6 \\ &= x(x+3)+2(x+3) \\ &= (x+2)(x+3)\end{aligned}\]Step 4: Combine the factors
Factorization by grouping
Let \(f(x)=x^3-3x^2-4x+12\). Group the first two terms and the last two terms:
\[\begin{aligned}f(x) &= (x^3-3x^2)-(4x-12) \\ &= x^2(x-3)-4(x-3) \\ &= (x-3)(x^2-4)\end{aligned}\]Use \(a^2-b^2=(a-b)(a+b)\) for \(x^2-4\):
\[(x-3)(x^2-4)=(x-3)(x-2)(x+2)\]Step 1: Find a zero
Let \(f(x)=x^3-8x^2+19x-12\).
\[f(1)=1-8+19-12=0\]So \((x-1)\) is a factor.
Step 2: Synthetic division by \(1\)
\[\begin{array}{r|rrrr} 1 & 1 & -8 & 19 & -12 \\ & & 1 & -7 & 12 \\ \hline & 1 & -7 & 12 & 0 \end{array}\]The quotient is \(q(x)=x^2-7x+12\).
Step 3: Factorize the quotient
\[\begin{aligned}x^2-7x+12 &= x^2-4x-3x+12 \\ &= x(x-4)-3(x-4) \\ &= (x-3)(x-4)\end{aligned}\]Step 4: Combine the factors
Step 1: Expand the expression
\[\begin{aligned}(x+2)(x^2-11x+48)-120 &= x^3-11x^2+48x+2x^2-22x+96-120 \\ &= x^3-9x^2+26x-24\end{aligned}\]Step 2: Find a zero
Let \(f(x)=x^3-9x^2+26x-24\).
\[f(2)=8-36+52-24=0\]So \((x-2)\) is a factor.
Step 3: Synthetic division by \(2\)
\[\begin{array}{r|rrrr} 2 & 1 & -9 & 26 & -24 \\ & & 2 & -14 & 24 \\ \hline & 1 & -7 & 12 & 0 \end{array}\]The quotient is \(q(x)=x^2-7x+12=(x-3)(x-4)\).
Step 4: Combine the factors
Note on the expression
The printed term \(3x^2-x^2-10x+2\) appears to be a misprint of \(3x^2-10x+2\), the same expression used in Question 4(f). The solution below uses \((x-3)(3x^2-10x+2)-10\).
Step 1: Expand the expression
\[\begin{aligned}(x-3)(3x^2-10x+2)-10 &= 3x^3-10x^2+2x-9x^2+30x-6-10 \\ &= 3x^3-19x^2+32x-16\end{aligned}\]Step 2: Find a zero
Let \(f(x)=3x^3-19x^2+32x-16\).
\[f(1)=3-19+32-16=0\]So \((x-1)\) is a factor.
Step 3: Synthetic division by \(1\)
\[\begin{array}{r|rrrr} 1 & 3 & -19 & 32 & -16 \\ & & 3 & -16 & 16 \\ \hline & 3 & -16 & 16 & 0 \end{array}\]The quotient is \(3x^2-16x+16\).
Step 4: Factorize the quotient
\[\begin{aligned}3x^2-16x+16 &= 3x^2-12x-4x+16 \\ &= 3x(x-4)-4(x-4) \\ &= (x-4)(3x-4)\end{aligned}\]Step 5: Combine the factors
Q.4 Solve the equations:
- a. \(x^3-3x^2-10x+24=0\)
- b. \(2x^3+3x^2-11x-6=0\)
- c. \(x^3+26x=9x^2+24\)
- d. \(x^3-4x^2=17x-60\)
- e. \((x+4)(x^2-x-14)+16=0\)
- f. \((x-3)(3x^2-10x+2)=10\)
(समीकरण हल गर्नुहोस् : )
Step 1: Find the first root
Let \(f(x)=x^3-3x^2-10x+24\).
\[f(2)=8-12-20+24=0\]So \(x=2\) is a root and \((x-2)\) is a factor.
Step 2: Synthetic division by \(2\)
\[\begin{array}{r|rrrr} 2 & 1 & -3 & -10 & 24 \\ & & 2 & -2 & -24 \\ \hline & 1 & -1 & -12 & 0 \end{array}\]The quotient is \(x^2-x-12=(x-4)(x+3)\).
Step 3: Solve for the roots
\[(x-2)(x-4)(x+3)=0 \implies x=2,\ x=4,\ x=-3\]Step 1: Find the first root
Let \(f(x)=2x^3+3x^2-11x-6\).
\[f(2)=16+12-22-6=0\]So \(x=2\) is a root.
Step 2: Synthetic division by \(2\)
\[\begin{array}{r|rrrr} 2 & 2 & 3 & -11 & -6 \\ & & 4 & 14 & 6 \\ \hline & 2 & 7 & 3 & 0 \end{array}\]The quotient is \(2x^2+7x+3\).
Step 3: Factorize the quotient
\[\begin{aligned}2x^2+7x+3 &= 2x^2+6x+x+3 \\ &= 2x(x+3)+1(x+3) \\ &= (x+3)(2x+1)\end{aligned}\]Step 4: Solve for the roots
\[(x-2)(x+3)(2x+1)=0 \implies x=2,\ x=-3,\ x=-\frac{1}{2}\]Step 1: Rearrange to standard form
\[x^3-9x^2+26x-24=0\]Step 2: Find the first root
Let \(f(x)=x^3-9x^2+26x-24\).
\[f(2)=8-36+52-24=0\]So \(x=2\) is a root.
Step 3: Synthetic division by \(2\)
\[\begin{array}{r|rrrr} 2 & 1 & -9 & 26 & -24 \\ & & 2 & -14 & 24 \\ \hline & 1 & -7 & 12 & 0 \end{array}\]The quotient is \(x^2-7x+12=(x-3)(x-4)\).
Step 4: Solve for the roots
\[(x-2)(x-3)(x-4)=0 \implies x=2,\ x=3,\ x=4\]Step 1: Rearrange
\[x^3-4x^2-17x+60=0\]Step 2: Find the first root
Let \(f(x)=x^3-4x^2-17x+60\).
\[f(3)=27-36-51+60=0\]So \(x=3\) is a root.
Step 3: Synthetic division by \(3\)
\[\begin{array}{r|rrrr} 3 & 1 & -4 & -17 & 60 \\ & & 3 & -3 & -60 \\ \hline & 1 & -1 & -20 & 0 \end{array}\]The quotient is \(x^2-x-20=(x-5)(x+4)\).
Step 4: Solve for the roots
\[(x-3)(x-5)(x+4)=0 \implies x=3,\ x=5,\ x=-4\]Step 1: Expand and simplify
\[\begin{aligned}x(x^2-x-14)+4(x^2-x-14)+16 &= x^3-x^2-14x+4x^2-4x-56+16 \\ &= x^3+3x^2-18x-40\end{aligned}\]So the equation is \(x^3+3x^2-18x-40=0\).
Step 2: Find the first root
Let \(f(x)=x^3+3x^2-18x-40\).
\[f(-2)=-8+12+36-40=0\]So \(x=-2\) is a root.
Step 3: Synthetic division by \(-2\)
\[\begin{array}{r|rrrr} -2 & 1 & 3 & -18 & -40 \\ & & -2 & -2 & 40 \\ \hline & 1 & 1 & -20 & 0 \end{array}\]The quotient is \(x^2+x-20=(x+5)(x-4)\).
Step 4: Solve for the roots
\[(x+2)(x+5)(x-4)=0 \implies x=-2,\ x=-5,\ x=4\]Step 1: Expand and simplify
\[\begin{aligned}& x(3x^2-10x+2)-3(3x^2-10x+2)-10=0 \\ &\implies 3x^3-10x^2+2x-9x^2+30x-6-10=0 \\ &\implies 3x^3-19x^2+32x-16=0\end{aligned}\]Step 2: Find the first root
Let \(f(x)=3x^3-19x^2+32x-16\).
\[f(1)=3-19+32-16=0\]So \(x=1\) is a root.
Step 3: Synthetic division by \(1\)
\[\begin{array}{r|rrrr} 1 & 3 & -19 & 32 & -16 \\ & & 3 & -16 & 16 \\ \hline & 3 & -16 & 16 & 0 \end{array}\]The quotient is \(3x^2-16x+16=(x-4)(3x-4)\).
Step 4: Solve for the roots
\[(x-1)(x-4)(3x-4)=0 \implies x=1,\ x=4,\ x=\frac{4}{3}\]Frequently Asked Questions on Class 10 Optional Mathematics Polynomial
What is covered in Class 10 Optional Mathematics Polynomial (Unit 2)? (कक्षा १० ऐच्छिक गणित बहुपदीय (एकाइ २) मा के के समावेश छ ?)
Class 10 Optional Mathematics Polynomial is Unit 2 of the Grade 10 Optional Mathematics book of Nepal. It has three exercises: Exercise 2.1 Remainder Theorem, Exercise 2.2 Factor Theorem and Exercise 2.3 Factorization and solving polynomial equations. This page solves all 45 questions.
What is the remainder theorem? (शेष साध्य भनेको के हो ?)
If a polynomial \(f(x)\) is divided by \((x-a)\), the remainder is \(f(a)\). For example, when \(x^3-x^2+1\) is divided by \(x-2\), the remainder is \(f(2)=8-4+1=5\) (Exercise 2.1, Question 3a).
How do I find the remainder when a polynomial is divided by \(ax+b\)? (बहुपदीयलाई \(ax+b\) ले भाग गर्दा शेष कसरी निकाल्ने ?)
Put \(ax+b=0\) to get \(x=-\dfrac{b}{a}\), then the remainder is \(f\left(-\dfrac{b}{a}\right)\). For example, for the divisor \(3x-4\) use \(x=\dfrac{4}{3}\) (Exercise 2.1, Question 3c).
What is the factor theorem? (गुणनखण्ड साध्य भनेको के हो ?)
A linear polynomial \((x-a)\) is a factor of \(f(x)\) if and only if \(f(a)=0\). In other words, the remainder must be zero.
What is the difference between the remainder theorem and the factor theorem? (शेष साध्य र गुणनखण्ड साध्यबीच के फरक छ ?)
The remainder theorem gives the remainder \(f(a)\) for any divisor \((x-a)\). The factor theorem is the special case where \(f(a)=0\), which means \((x-a)\) divides \(f(x)\) exactly.
How do I find an unknown constant such as \(k\) using these theorems? (यी साध्यहरू प्रयोग गरी \(k\) जस्तो अज्ञात स्थिरांक कसरी निकाल्ने ?)
- Find the value of \(x\) that makes the divisor zero.
- Substitute it in \(f(x)\) and equate to the given remainder (or to \(0\) for a factor).
- Solve the equation for \(k\).
Example: Exercise 2.2, Question 5a gives \(k=12\).
How do I find the number to be added or subtracted so that \((x-a)\) becomes a factor? (\((x-a)\) गुणनखण्ड हुन कति जोड्नु वा घटाउनुपर्छ भनी कसरी पत्ता लगाउने ?)
Let the number be \(k\), form \(g(x)=f(x)+k\) (or \(f(x)-k\)), then solve \(g(a)=0\). See Exercise 2.2, Question 6.
How do I factorize a cubic polynomial? (घन बहुपदीयलाई कसरी खण्डीकरण गर्ने ?)
- Try values such as \(\pm1,\pm2,\pm3\) (factors of the constant term) until \(f(a)=0\).
- Divide by \((x-a)\) using synthetic division.
- Factorize the quadratic quotient.
- Write all the factors together.
What is synthetic division? (सिन्थेटिक डिभिजन भनेको के हो ?)
It is a short method of dividing a polynomial by \((x-a)\) using only the coefficients. Write the coefficients, bring down the first one, multiply by \(a\), add to the next coefficient, and repeat. The last number is the remainder, and the other numbers are the coefficients of the quotient.
How do I solve a cubic equation such as \(x^3-3x^2-10x+24=0\)? (\(x^3-3x^2-10x+24=0\) जस्तो घन समीकरण कसरी हल गर्ने ?)
Find one root by trial, for example \(f(2)=0\). Divide by \((x-2)\) to get a quadratic, factorize it, and set every factor equal to zero. The roots are \(2,\,4,\,-3\) (Exercise 2.3, Question 4a).
Are these questions useful for SEE preparation in Nepal? (के यी प्रश्नहरू नेपालमा SEE को तयारीका लागि उपयोगी छन् ?)
Yes, for students who study Optional Mathematics. This set covers every question type in Exercises 2.1, 2.2 and 2.3: proofs, finding remainders, unknown constants, factorization and solving cubic equations. Practising them builds the skills needed for SEE-style questions.
Are the solutions available in Nepali? (के समाधान नेपाली भाषामा पनि उपलब्ध छन् ?)
Every question is shown in English with the Nepali question in brackets. Multiple choice solutions, definitions and proofs also have Nepali in brackets. All other solutions are in English only, because the steps and symbols of mathematics are the same in both languages.
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