Factorise the expressions and divide them as directed:
(i) (x² - 22x + 117) ÷ (x - 13)
(ii) (x³ + x² - 132x) ÷ x (x - 11)
(iii) (2x³ - 12x² + 16x) ÷ (x - 2) (x - 4)
(iv) (9x² - 4) ÷ (3x + 2)
(v) (3x² - 48) ÷ (x - 4)
(vi) (x⁴ - 16) ÷ x³ + 2x² + 4x + 8
(vii) (3x⁴ - 1875) ÷ (3x² - 75)
Solution:
(i) (x² - 22x + 117) ÷ (x - 13)
Given, (x² - 22x + 117) ÷ (x - 13)
Factorising the numerator,
(x² - 22x + 117) = x² - 13x - 9x + 117
= x(x -13) - 9(x - 13)
= (x - 13) (x - 9)
Now, (x² - 22x + 117) ÷ (x - 13)
= [(x - 13) (x - 9)] ÷ (x - 13)
= (x - 9)
(ii) (x³ + x² - 132x) ÷ x (x - 11)
Given, (x³ + x² - 132x) ÷ x (x - 11)
Factorising the numerator,
(x³ + x² - 132x) = x[x² + x - 132]
= x[x² + 12x - 11x - 132]
= x[x(x + 12) - 11(x + 12)]
= x[(x + 12) (x - 11)]
Now, (x³ + x² - 132x) ÷ x (x - 11)
= x[(x + 12) (x - 11)] ÷ x (x - 11)
= (x +12)
(iii) (2x³ - 12x² + 16x) ÷ (x - 2) (x - 4)
Given, (2x³ - 12x² + 16x) ÷ (x - 2) (x - 4)
Factorising the numerator,
(2x³ - 12x² + 16x) = 2x[x² - 6x + 8]
= 2x[x² - 4x - 2x + 8]
= 2x[x(x - 4) - 2(x - 4)]
= 2x[(x - 4) (x - 2)]
Now, (2x³ - 12x² + 16x) ÷ (x - 2) (x - 4)
= 2x[(x - 4) (x - 2)] ÷ (x - 2) (x - 4)
= 2x
(iv) (9x² - 4) ÷ (3x + 2)
Given, (9x² - 4) ÷ (3x + 2)
Factorising the numerator,
(9x² - 4) = [(3x)² - 2²]
= (3x + 2)(3x - 2)
Now, (9x² - 4) ÷ (3x + 2) = (3x + 2)
= (3x + 2)(3x - 2) ÷ (3x + 2)
= (3x - 2)
(v) (3x² - 48) ÷ (x - 4)
Given, (3x² - 48) ÷ (x - 4)
Factorising the numerator,
(3x² - 48) = 3(x² - 16)
= 3[(x)² - (4)²]
= 3[(x + 4)(x - 4)]
Now, (3x² - 48) ÷ (x - 4)
= 3[(x + 4)(x - 4)] ÷ (x - 4)
= 3(x + 4)
(vi) (x⁴ - 16) ÷ x³ + 2x² + 4x + 8
Given, (x⁴ - 16) ÷ x³ + 2x² + 4x + 8
Factorising the numerator,
(x⁴ - 16) = [(x²)² - (4)²]
= [(x² + 4)(x² - 4)]
= [(x² + 4)((x)² - (2)²)]
= (x² + 4) (x + 2)(x - 2)
Factorising the denominator,
x³ + 2x² + 4x + 8 = x²(x + 2) + 4(x + 2)
= (x + 2)(x² + 4)
Now, (x⁴ - 16) ÷ x³ + 2x² + 4x + 8
= [(x² + 4) (x + 2)(x - 2)] ÷ [(x + 2)(x² + 4)]
= (x - 2)
(vii) (3x⁴ - 1875) ÷ (3x² - 75)
Given, (3x⁴ - 1875) ÷ (3x² - 75)
Factorising the numerator,
(3x⁴ - 1875) = 3(x⁴ - 625) = 3[(x²)² - (25)²]
= 3[(x² + 25) (x² - 25)]
= 3(x² + 25) [(x)² - (5)²]
= 3(x² + 25) (x + 5) (x - 5)
Factoring the denominator,
(3x² - 75) = 3(x² - 25)
= 3 [(x)² - (5)²]
= 3 [(x + 5) (x - 5)]
Now, (3x⁴ - 1875) ÷ (3x² - 75)
= [3(x² + 25) (x + 5) (x - 5)] ÷ [3(x + 5) (x - 5)]
= (x² + 25)
✦ Try This: Factorise the expressions and divide them as directed:
(i) (x² - 12x - 85) ÷ ( x + 5), (ii) (x³ - 4x)(x² + x - 12) ÷ (x - 2)(x - 3)
☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 9
NCERT Exemplar Class 8 Maths Chapter 7 Problem 96
Factorise the expressions and divide them as directed: (i) (x² - 22x + 117) ÷ (x - 13), (ii) (x³ + x² - 132x) ÷ x (x - 11), (iii) (2x³ - 12x² + 16x) ÷ (x - 2) (x - 4), (iv) (9x² - 4) ÷ (3x + 2), (v) (3x² - 48) ÷ (x - 4), (vi) (x⁴ - 16) ÷ x³ + 2x² + 4x + 8, (vii) (3x⁴ - 1875) ÷ (3x² - 75)
Summary:
Factorising and dividing the expressions (i) (x² - 22x + 117) ÷ (x - 13), (ii) (x³ + x² - 132x) ÷ x (x - 11), (iii) (2x³ - 12x² + 16x) ÷ (x - 2) (x - 4), (iv) (9x² - 4) ÷ (3x + 2), (v) (3x² - 48) ÷ (x - 4), (vi) (x⁴ - 16) ÷ x³ + 2x² + 4x + 8, (vii) (3x⁴ - 1875) ÷ (3x² - 75) we get, (x - 9), (x +12), 2x, (3x - 2), 3(x + 4), (x - 2), (x² + 25) respectively
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