Question

Factorise
(i) p² -25q²

(ii) 2x²+13x+20

(iii) 4x²+9y²+16z²+12xy-24yz-16xz

All Steps Required *

Answer 1

Answer:

(i) p²-25q²=(p)²-(5q)²=(p+5q)(p-5p). a²-b²=(a+b)(a-b)

(ii) 2x²+13x+20=2x²+(8+5)x+20

=2x²+8x+5x+20

=2x(x+4)+5(x+4)=(x+4)(2x+5)

(iii) 4x²+9y²+16z²+12xy-24yz-16xz

=(2x)²+(3y)²+(-4z)²+2×2x×3y+2×3y(-4z)+2×2x×(-4z)

=(2x+3y-4z)²

=(2x+3y-4z)(2x+3y-4z)

Answer 2

$\huge\sf\underline\red{Answer}$

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$\small\purple{\textbf {\textsf {p² -25q²}}}$

$p {}^{2} - 25q {}^{2} \\ = (p) {}^{2} - (5q) {}^{2} \\ = ( p + 5q)(p - 5q)$

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$\small\purple{\textbf {\textsf {2x²+13x+20}}}$

$2x {}^{2} + 13x + 20 \\ = 2x {}^{2} + 8x + 5x + 20 \\ = 2x(x + 4) + 5(x + 4) \\ = (x + 4)(2x + 5)$

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$\small\purple{\textbf {\textsf {4x²+9y²+16z²+12xy-24yz-16xz}}}$

$4x²+9y²+16z²+12xy-24yz-16xz \\ \\ = (2x) {}^{2} + (3y) {}^{2} + (4z) {}^{2} + 2(2)(3) xy\\ - 2(3)(4)yz - 2(2x)(4y) \\ \\ = (2x + 3y - 4z)( 2x + 3y - 4z)$

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