Question

factorise it
a)(x^2+7x+12)+(x+3)
b)(x-3)^2 by 4x^2-25 + (x^2-x-6) by 2x-5

Answer 1

Hey there!

a) Factorisation:

$\bf{(x^2 + 7x + 12) + (x + 3)}$

$\bf{(x^2 + 3x) + (4x + 12) + (x + 3)}$

$\bf{x(x + 3) + 4(x + 3) + (x + 3)}$

$\bf{x + (x + 3) (x + 4) + 3}$

$\bf{(x + 3) (x + 4 + 1)}$

$\boxed{\bf{\underline{\therefore \quad (x^2 + 7x + 12) + (x + 3) = (x + 3)(x + 5)}}}$

b) If told for a conjoint factorisation, then this is the first part of the solution:

$\bf{\dfrac{(x - 3)^2}{(2x + 5) (2x + 5)} + \dfrac{x^2 - x - 6}{2x - 5}}$

$\bf{\dfrac{(x - 3)^2}{(2x + 5) (2x - 5)} + \dfrac{(x + 2)(x - 3)}{2x - 5}}$

$\bf{\dfrac{(x - 3)^2}{(2x + 5) (2x - 5)} + \dfrac{(x + 2)(x - 3)(2x + 5)}{(2x - 5)(2x + 5)}}$

$\bf{\dfrac{(x - 3)^2 + (x + 2)(x - 3)(2x + 5)}{(2x + 5)(2x - 5)}}$

$\bf{\dfrac{x^2 - 6x + 9 + (x + 2)(x - 3)(2x + 5)}{(2x + 5)(2x - 5)}}$

$\bf{\dfrac{x^2 - 6x + 9 + 2x^3 + 3x^2 - 17x - 30}{(2x + 5)(2x - 5)}}$

$\bf{\dfrac{2x^3 + 4x^2 - 23x + 9 - 30}{(2x + 5)(2x - 5)}}$

$\bf{\dfrac{2x^3 + 4x^2 - 23x - 21}{(2x + 5)(2x - 5)}}$

$\boxed{\bf{\underline{Final \: \: Answer: \: \: \dfrac{(x - 3) \Bigg(2x^2 + 10x + 7 \Bigg)}{(2x + 5)(2x - 5)}}}}$

Hope it helps and clears the doubts for factorising the products!!!