Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.8

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.8

 

Question 1. Factorise each of the following polynomials using synthetic division:
(i) \( x^3 – 3x^2 – 10x + 24 \)
Answer:We define the polynomial as \( p(x) = x^3 – 3x^2 – 10x + 24 \). First, we test if \( x-1 \) is a factor by substituting \( x=1 \) into the polynomial:
\( p(1) = (1)^3 - 3(1)^2 - 10(1) + 24 \)
\( = 1 - 3 - 10 + 24 \)
\( = 25 - 13 \)
\( = 12 \)
Since \( p(1) \neq 0 \), \( x-1 \) is not a factor. Next, we test if \( x+1 \) is a factor by substituting \( x=-1 \):
\( p(-1) = (-1)^3 - 3(-1)^2 - 10(-1) + 24 \)
\( = -1 - 3(1) + 10 + 24 \)
\( = -1 - 3 + 10 + 24 \)
\( = 34 - 4 \)
\( = 30 \)
Since \( p(-1) \neq 0 \), \( x+1 \) is not a factor. Now, we test if \( x-2 \) is a factor by substituting \( x=2 \):
\( p(2) = (2)^3 - 3(2)^2 - 10(2) + 24 \)
\( = 8 - 3(4) - 20 + 24 \)
\( = 8 - 12 - 20 + 24 \)
\( = 32 - 32 \)
\( = 0 \)
Since \( p(2) = 0 \), \( x-2 \) is a factor of the polynomial. This means we can proceed with synthetic division using 2. We perform synthetic division with 2 using the coefficients of \( p(x) \):

The resulting quotient is \( x^2 - x - 12 \). Now, we factorise this quadratic expression:
\( x^2 - x - 12 = x^2 - 4x + 3x - 12 \)
\( = x(x - 4) + 3(x - 4) \)
\( = (x - 4)(x + 3) \) So, the factors of the polynomial \( x^3 - 3x^2 - 10x + 24 \) are \( (x - 2)(x - 4)(x + 3) \).In simple words: We find a number that makes the polynomial equal to zero when plugged in. Then, we use synthetic division with that number to reduce the polynomial to a simpler quadratic expression. Finally, we factor the quadratic expression to find all the polynomial's factors.

🎯 Exam Tip: Always start testing simple integer values like \( \pm 1, \pm 2, \pm 3 \) using the Rational Root Theorem to find the first factor quickly. This simplifies the polynomial for further factorization.

 

Question 1. Factorise each of the following polynomials using synthetic division:
(ii) \( 2x^3 – 3x^2 – 3x + 2 \)
Answer:We define the polynomial as \( p(x) = 2x^3 – 3x^2 – 3x + 2 \). First, we test if \( x-1 \) is a factor by substituting \( x=1 \):
\( p(1) = 2(1)^3 - 3(1)^2 - 3(1) + 2 \)
\( = 2 - 3 - 3 + 2 \)
\( = 4 - 6 \)
\( = -2 \)
Since \( p(1) \neq 0 \), \( x-1 \) is not a factor. Next, we test if \( x+1 \) is a factor by substituting \( x=-1 \):
\( p(-1) = 2(-1)^3 - 3(-1)^2 - 3(-1) + 2 \)
\( = 2(-1) - 3(1) + 3 + 2 \)
\( = -2 - 3 + 3 + 2 \)
\( = 5 - 5 \)
\( = 0 \)
Since \( p(-1) = 0 \), \( x+1 \) is a factor of the polynomial. This means we can perform synthetic division with -1. We perform synthetic division with -1 using the coefficients of \( p(x) \):

The resulting quotient is \( 2x^2 - 5x + 2 \). Now, we factorise this quadratic expression:
\( 2x^2 - 5x + 2 = 2x^2 - 4x - x + 2 \)
\( = 2x(x - 2) - 1(x - 2) \)
\( = (x - 2)(2x - 1) \) So, the factors of the polynomial \( 2x^3 - 3x^2 - 3x + 2 \) are \( (x + 1)(x - 2)(2x - 1) \).In simple words: First, find a value for \( x \) that makes the polynomial equal to zero. This value helps identify one factor. Then, use synthetic division to break the polynomial down into a quadratic part. Finally, factorise this quadratic part to find the remaining factors.

🎯 Exam Tip: Remember that for a polynomial to have \( (x-a) \) as a factor, substituting \( x=a \) into the polynomial must result in zero (Remainder Theorem).

 

Question 1. Factorise each of the following polynomials using synthetic division:
(iii) \( -7x + 3 + 4x^3 \)
Answer:First, we arrange the polynomial in standard form: \( p(x) = 4x^3 + 0x^2 - 7x + 3 \). We test if \( x-1 \) is a factor by substituting \( x=1 \):
\( p(1) = 4(1)^3 - 7(1) + 3 \)
\( = 4 - 7 + 3 \)
\( = 7 - 7 \)
\( = 0 \)
Since \( p(1) = 0 \), \( x-1 \) is a factor of the polynomial. This means we can perform synthetic division with 1. We perform synthetic division with 1 using the coefficients of \( p(x) \):

The resulting quotient is \( 4x^2 + 4x - 3 \). Now, we factorise this quadratic expression:
\( 4x^2 + 4x - 3 = 4x^2 + 6x - 2x - 3 \)
\( = 2x(2x + 3) - 1(2x + 3) \)
\( = (2x + 3)(2x - 1) \) So, the factors of the polynomial \( -7x + 3 + 4x^3 \) are \( (x - 1)(2x + 3)(2x - 1) \).In simple words: First, put the polynomial in the correct order from highest power to lowest. Find a value that makes the polynomial zero. Use this value for synthetic division to get a quadratic part, then factorise that quadratic part to find all factors.

🎯 Exam Tip: When a term like \( x^2 \) is missing from the polynomial, remember to use \( 0 \) as its coefficient in synthetic division to keep the place value correct.

 

Question 1. Factorise each of the following polynomials using synthetic division:
(iv) \( x^3 + x^2 – 14x – 24 \)
Answer:We define the polynomial as \( p(x) = x^3 + x^2 – 14x – 24 \). First, we test if \( x-1 \) is a factor by substituting \( x=1 \):
\( p(1) = (1)^3 + (1)^2 - 14(1) - 24 \)
\( = 1 + 1 - 14 - 24 \)
\( = 2 - 38 \)
\( = -36 \)
Since \( p(1) \neq 0 \), \( x-1 \) is not a factor. Next, we test if \( x+1 \) is a factor by substituting \( x=-1 \):
\( p(-1) = (-1)^3 + (-1)^2 - 14(-1) - 24 \)
\( = -1 + 1 + 14 - 24 \)
\( = 15 - 25 \)
\( = -10 \)
Since \( p(-1) \neq 0 \), \( x+1 \) is not a factor. Now, we test if \( x+2 \) is a factor by substituting \( x=-2 \):
\( p(-2) = (-2)^3 + (-2)^2 - 14(-2) - 24 \)
\( = -8 + 4 + 28 - 24 \)
\( = 32 - 32 \)
\( = 0 \)
Since \( p(-2) = 0 \), \( x+2 \) is a factor of the polynomial. This means we can perform synthetic division with -2. We perform synthetic division with -2 using the coefficients of \( p(x) \):

The resulting quotient is \( x^2 - x - 12 \). Now, we factorise this quadratic expression:
\( x^2 - x - 12 = x^2 - 4x + 3x - 12 \)
\( = x(x - 4) + 3(x - 4) \)
\( = (x - 4)(x + 3) \) So, the factors of the polynomial \( x^3 + x^2 - 14x - 24 \) are \( (x + 2)(x - 4)(x + 3) \).In simple words: When a polynomial is given, first check different simple numbers to find one that makes the polynomial zero. This gives you one factor. Then, use synthetic division to get a simpler quadratic equation. Factor this quadratic equation to find the rest of the factors.

🎯 Exam Tip: Be systematic when testing potential factors, starting with small integers, and remember that if \( p(a) = 0 \), then \( (x-a) \) is a factor.

 

Question 1. Factorise each of the following polynomials using synthetic division:
(v) \( x^3 – 7x + 6 \)
Answer:We define the polynomial as \( p(x) = x^3 + 0x^2 - 7x + 6 \). We test if \( x-1 \) is a factor by substituting \( x=1 \):
\( p(1) = (1)^3 - 7(1) + 6 \)
\( = 1 - 7 + 6 \)
\( = 7 - 7 \)
\( = 0 \)
Since \( p(1) = 0 \), \( x-1 \) is a factor of the polynomial. This means we can perform synthetic division with 1. We perform synthetic division with 1 using the coefficients of \( p(x) \):

The resulting quotient is \( x^2 + x - 6 \). Now, we factorise this quadratic expression:
\( x^2 + x - 6 = x^2 + 3x - 2x - 6 \)
\( = x(x + 3) - 2(x + 3) \)
\( = (x + 3)(x - 2) \) So, the factors of the polynomial \( x^3 - 7x + 6 \) are \( (x - 1)(x - 2)(x + 3) \).In simple words: When a polynomial is given, check for simple numbers that make it zero to find one factor. Use synthetic division to reduce it to a quadratic expression. Then, factor this quadratic expression to find the remaining factors.

🎯 Exam Tip: For polynomials with missing terms, always include a zero coefficient for that term in synthetic division to ensure correct alignment of terms.

 

Question 1. Factorise each of the following polynomials using synthetic division:
(vi) \( x^3 – 10x^2 – x + 10 \)
Answer:We define the polynomial as \( p(x) = x^3 – 10x^2 – x + 10 \). We test if \( x-1 \) is a factor by substituting \( x=1 \):
\( p(1) = (1)^3 - 10(1)^2 - 1(1) + 10 \)
\( = 1 - 10 - 1 + 10 \)
\( = 11 - 11 \)
\( = 0 \)
Since \( p(1) = 0 \), \( x-1 \) is a factor of the polynomial. This means we can perform synthetic division with 1. We perform synthetic division with 1 using the coefficients of \( p(x) \):

The resulting quotient is \( x^2 - 9x - 10 \). Now, we factorise this quadratic expression:
\( x^2 - 9x - 10 = x^2 - 10x + x - 10 \)
\( = x(x - 10) + 1(x - 10) \)
\( = (x - 10)(x + 1) \) So, the factors of the polynomial \( x^3 - 10x^2 - x + 10 \) are \( (x - 1)(x - 10)(x + 1) \).In simple words: First, find a number that makes the polynomial's value zero. Use this number in synthetic division to get a simpler quadratic expression. Then, factorise this quadratic expression to find the remaining factors.

🎯 Exam Tip: When factorising a quadratic expression \( ax^2 + bx + c \), look for two numbers that multiply to \( ac \) and add up to \( b \). This helps break down the middle term.