## ML Aggarwal Class 8 Factorisation Check Your Progress

Question 1.

Find the HCF of the given polynomials:

(i) 14pq, 28p^{2}q^{2}

(ii) 8abc, 24ab^{2}, 12a^{2}b

Solution:

(i) 14pq, 28p^{2}q^{2} (HCF of 14, 28 = 14)

HCF of 1 4pq, 28p^{2}q^{2} = 14pq

(ii) 8abc, 24ab^{2}, 12a^{2}b

HCF of 8, 24, 12 = 4

HCF of 8abc, 24ab^{2}, 12a^{2}b = 4ab

Question 2.

Factorise the following:

(i) 10x^{2} – 18x^{3} + 14x^{4}

(ii) 5x^{2}y + 10xyz + 15xy^{2}

(iii) p^{2}x^{2} + c^{2}x^{2} – ac^{2} – ap^{2}

(iv) 15(x + y)^{2} – 5x – 5y

(v) (ax + by)^{2} + (ay – bx)^{2}

(vi) ax + by + cx + bx + cy + ay

(vii) 49x^{2} – 70xy + 25y^{2}

(viii) 4a^{2} + 12ab + 9b^{2}

(ix) 49p^{2} – 36q^{2}

(x) 100x^{3} – 25xy^{2}

(xi) x^{2} – 2xy + y^{2} – z^{2}

(xii) x^{8} – y^{8}

(xiii) 12x^{3} – 14x^{2} – 10x

(xiv) p^{2} – 10p + 21

(xv) 2x^{2} – x – 6

(xvi) 6x^{2} – 5xy – 6y^{2}

(xvii) x^{2} + 2xy – 99y^{2}

Solution:

(i) 10x^{2} – 18x^{3} + 14x^{4}

HCF of 10, 18, 14 = 2

∴ 10x^{2} – 18x^{3/sup> + 14 x 4 = 2x2 (5 – 9x + 7x2)}

(ii) 5x^{2}y + 10xyz + 15xy^{2}

HCF of 5, 10, 15 = 5

∴ 5x^{2}y + 10xyz + 15xy^{2}

= 5xy(x + 2z + 3y)

(iii) p^{2}x^{2} + c^{2}x^{2} – ac^{2} – ap

= p^{2}x^{2} – ap^{2} + c^{2}x^{2} – ac^{2}

p^{2}(x^{2} – a) + c^{2}(x^{2} – a)

= (x^{2} – a) (p^{2} + c^{2})

(iv) 15(x + y)^{2} – 5x – 5y

= 15(x + y)^{2} – 5(x + y)

= 5(x + y) [3(x + y) – 1]

= 5(x + y) (3x + 3y – 1)

(v) (ax + by)^{2} + (ay – bx)^{2}

a^{2}x^{2} + b^{2}y^{2} + 2abxy + a^{2}y^{2} + b^{2}x^{2} – 2abxy

= a^{2}x^{2} + a^{2}y^{2} + b^{2}x^{2} + b^{2}y^{2}

= a^{2}(x^{2} + y^{2}) + b^{2}(x^{2} + y^{2})

= (x^{2} + y^{2}) (a^{2} + b^{2})

(vi) ax + by + cx + bx + cy + ay

= ax + bx + cx + ay + by + cy (grouping)

= x(a + b + c) + y(a + b + c)

= (a + b + c) (x + y)

(vii) 49x^{2} – 70xy + 25y^{2}

= (7x)^{2} – 2 × 7x × 5y + (5y)^{2}

{∵ (a – b)^{2} = a^{2} – 2ab + b^{2}}

= (7x – 5y)^{2}

(viii) 4a^{2} + 12ab + 9b^{2}

= (2a)2^{2} + 2 × 2a × 3b + (3b)^{2}

{∵ (a + b)^{2} = a^{2} + 2ab + b^{2}}

= (2a + 3b)^{2}

(ix) 49p^{2} – 36q^{2}

= (7p)^{2} – (6q)^{2}

= (7p + 6q) (7p – 6q)

{∵ a^{2} – b^{2} = (a + b) (a – b)}

(x) 100x^{3} – 25xy^{2}

= 25x(x^{2} – y^{2}) = 25x{(x)^{2} – (y)^{2}}

= 25x(x + y) (x – y)

(xi) x^{2} – 2xy + y^{2} – z^{2}

= (x – y)^{2} – (z)^{2}

{∵ a^{2} -2ab + b^{2} = (a – b)^{2})

a^{2} – b^{2} = (a + b)(a – b)}

= (x – y + z)(x – y – z)

(xii) x^{8} – y^{8}

= (x^{4})^{2} – (y^{4})^{2}

{∵ a^{2} – b^{2} = (a + b)(a- b)}

= (x^{4} + y^{4})(x^{4} – y^{4})

= (x^{4} + y^{4}) {(x^{2})^{2} – (y^{2})^{1}}

= (x^{4} + y^{4}) (x^{2} + y^{2}) (x^{2} – y^{2})

= (x^{4} + y^{4} (x^{2} + y^{2}) (x + y) (x – y)

(xiii) 12x^{3} – 14x^{2} – 10x

= 2x(6x^{2} – 7x – 5)

{∵ 6 × (-5) = -30

∴ -30 = -10 × 3

-7 = -10 + 3}

= 2x{6x^{2} + 3x – 10x – 5}

= 2x{3x(2x + 1) – 5(2x + 1)}

= 2x(2x + 1) (3x – 5)

(xiv) p^{2} – 10p + 21

= p^{2} – 3p – 7p + 21

{∵ 21 =-3 × (-7)

-10 = -3 – 7}

= p(p – 3) – 7(p – 3)

= (p – 3)(p – 7)

(xv) 2x^{2} – x – 6

= 2x^{2} – 4x + 3x – 6

{ ∵ -6 × 2 = -12

∴ -12 = -4 × 3

-1 = -4 + 3}

= 2x(x – 2) + 3(x – 2)

= (x – 2) (2x + 3)

(xvi) 6x^{2} – 5xy – 6y^{2}

= 6x^{2} – 9xy + 4xy – 6y^{2}

{∵ 6 × (-6) = -36

∴ – 36 = -9 × 4

– 5 = -9 + 4}

= 3x(2x – 3y) + 2y(2x – 3y)

= (2x – 3y) (3x + 2y)

(xvii) x^{2} + 2xy – 99y^{2}

= x^{2} + 11xy – 9xy – 99y^{2}

{∵ -99 = -11 × 9

-2 = -11 + 9 }

= x(x + 11y) – 9y(x + 11y)

= (x + 11y) (x – 9y)

Question 3.

Divide as directed:

(i) 15(y + 3)(y^{2} – 16) ÷ 5(y^{2} – y – 12)

(ii) (3x^{3} – 6x^{2} – 24x) ÷ (x – 4) (x + 2)

(iii) (x^{4} – 81) ÷ (x^{3} + 3x^{2} + 9x + 27)

Solution:

(i) 15(y + 3)(y^{2} – 16) ÷ 5(y^{2} – y – 12)

y^{2} – 16 = (y)^{2} – (4)^{2}

= (y + 4)(y – 4)

y^{2} – y – 12 = y^{2} – 4y + 3y – 12

= y(y – 4) + 3(y – 4)

= (y – 4)(y + 3)

Now, \(\frac{15(y+3)\left(y^{2}-16\right)}{5\left(y^{2}-y-12\right)}\)

\(\frac{15 \times(y+3)(y+4)(y-4)}{5(y-4)(y+3)}\)

= 3(y + 4)

(ii) (3x^{3} – 6x^{2} – 24x) ÷ (x – 4) (x + 2)

3x^{3} – 6x^{2} – 24x = 3x(x^{2} – 2x – 8)

= 3x{x^{2} – 4x + 2x – 8}

= 3x{x(x – 4) + 2(x – 4)}

= 3x(x – 4) (x + 2)

∴ \(\frac{3 x^{3}-6 x^{2}-24 x}{(x-4)(x+2)}=\frac{3 x(x-4)(x+2)}{(x-4)(x+2)}=3 x\)

(iii) (x^{4} – 81) ÷ (x^{3} + 3x^{2} + 9x + 27)

x^{4} – 81 = (x^{2})^{2} – (9)^{2} = (x^{2} + 9) (x^{2} – 9)

= (x^{2} + 9) {(x)^{2} – (3)^{2}}

= (x^{2} + 9) (x + 3) (x – 3)

x^{3} + 3x^{2} + 9x + 27

= (x)^{2} + (x + 3) + 9 (x + 3)

= (x^{2} + 9) (x + 3)

Now, \(\frac{x^{4}-81}{x^{3}+3 x^{2}+9 x+27}\)

= \(\frac{\left(x^{2}+9\right)(x+3)(x-3)}{\left(x^{2}+9\right)(x+3)}\)

= x = -3