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ICSE Class 10 Maths Chapter 8 Factorisation Solved

ICSE Class 10 Maths Chapter 8 Factorisation Solutions

ICSE Class 10 Maths Chapter 8 in Selina Concise Mathematics explains factorisation of polynomials through the Remainder Theorem and the Factor Theorem. For a divisor x-a, substitute x=a in f(x); the value f(a) is the remainder, and if f(a)=0, then x-a is a factor.

Concept snapshot

Think of x-a as a lock and a as its key. If substituting x=a gives 0, the lock opens: x-a is a factor. If it gives a non-zero value, that value is the remainder.

Remainder theorem and factor theorem method

TaskMethodConclusion
Remainder by x-aFind f(a)Remainder =f(a)
Factor check for x-aFind f(a)If f(a)=0, it is a factor
Divisor px+qUse x=-\frac{q}{p}Substitute this value in f(x)

Worked examples from Selina Chapter 8

Example 1: Find m if x-1 is a factor of 8x^2-7x+m

Step 1: Since x-1 is a factor, put x=1.

Step 2: Apply f(1)=0.

8(1)^2-7(1)+m=0

8-7+m=0 \Rightarrow 1+m=0 \Rightarrow m=-1

Final answer: m=-1.

Example 2: Factorise 2x^3+x^2-13x+6

Step 1: Test x=2.

2(2)^3+2^2-13(2)+6=16+4-26+6=0

Step 2: Hence x-2 is a factor. Divide and factorise the quotient.

2x^3+x^2-13x+6=(x-2)(2x^2+5x-3)

2x^2+5x-3=(2x-1)(x+3)

Final answer: (x-2)(2x-1)(x+3).

Example 3: Find p and q if px^3+4x^2-3x+q is divisible by x^2-1

Step 1: Since x^2-1=(x-1)(x+1), use x=1 and x=-1.

p+4-3+q=0 \Rightarrow p+q=-1

-p+4+3+q=0 \Rightarrow -p+q=-7

Step 2: Solve the equations.

p=3,\quad q=-4

3x^3+4x^2-3x-4=(x-1)(x+1)(3x+4)

Final answer: p=3, q=-4; factors are (x-1)(x+1)(3x+4).

Quick answer index for Chapter 8

ExerciseAnswers
8(A)1(a) m=-1; 1(b) -k; 1(c) k=6; 1(d) a=-5; 1(e) x^3-x^2+x-6; 2(i) 1; 2(ii) 0; 2(iii) 2; 3: x+1, x+2 are factors; 4(i) a=-5; 4(ii) k=8; 5 a=3,b=-4; 6 k=2; 7 a=\frac{3}{2}; 8 m=3,n=7; 9 k=\frac{20}{3}; 10 a=2; 11 a=-3,b=-1; 12 a=3,b=-3; 13 add -46; 14 subtract 8; 15 a=10; 16 a=5,b=-11; 17 a=3.
8(B)1(i) (x-2)(x-3)(x+3); 1(ii) (x+5)(2x+1)(x-3); 1(iii) (3x+2)(x-1)(x+1); 2(i) (x-2)(x+3)(3x-1); 2(ii) (x-2)(2x-1)(x+3); 2(iii) (x+2)(3x+5)(x-3); 2(iv) (x+3)(4x+7)(x-3); 2(v) (x+1)(x-2)(x+2); 3 roots -1,-3,\frac{2}{3}; 4 roots 2,3,-\frac{3}{2}; 5 a=-6,b=-8; 6 b=-8,c=3; 7 a=\frac{n-q}{m-p}; 8 a=1; 9 a=-2; 10 subtract 9.
8(C) / Test Yourself1 (x-1)(x-2)(x-4); 2 (x-1)(x+13)(x-2); 3 m=7; 4 subtract -67; 5 a=1,b=7; 6 a=-5,b=6; 7 (x+1)(x+2)(x+3); 8 m=\frac{3}{10}; 9 p=3,q=-4; 10 add -9; 11 a=4; 12 a=4, (3x+5)(x^2-3); 13 p=1; 14 (x-2)(2x-1)(x+3); 15 k=13; 16 subtract 1.

Examiner mindset for this chapter

Write the zero of the divisor before substitution. For 2x-3, the correct value is x=\frac{3}{2}, not x=3. For complete factorisation, do not stop after one factor; divide the cubic and factorise the quadratic quotient.

Common mistakes students make

  • Using the wrong sign for the root of a divisor such as 3x+2.
  • Writing f(x)+c when the question says a number is subtracted; it should be f(x)-c.
  • Giving only one factor when the question asks to factorise completely.
  • Forgetting that a zero remainder is the condition for a factor.

Continue revision with ICSE solutions by class and subject, ICSE Class 10 solutions, and Selina Concise Mathematics Class 10 solutions.

Frequently Asked Questions

How do I use the Remainder Theorem in ICSE Class 10 Maths?

For x-a, substitute x=a in f(x). The value f(a) is the remainder.

What does the Factor Theorem say?

The Factor Theorem says that x-a is a factor of f(x) if and only if f(a)=0.

How do I handle a divisor like 2x+1?

Solve 2x+1=0, so x=-\frac{1}{2}, and substitute that value in the polynomial.

What should I check before writing the final answer?

Check the substitution value, signs, simultaneous equations and whether the final factorisation is complete.

Sources and syllabus alignment

This page follows the ICSE Class 10 Mathematics treatment of polynomials and the Selina Concise Mathematics Chapter 8 structure.





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