Factorise (i) 7a 3 + 7a – 2a 2 – 2 (ii) 4ax + 3by – 3ay –4bx (iii) x 3 + y 3 + x 2 y + xy 2

Question

Factorise (i) 7a 3 + 7a &#x2013; 2a 2 &#x2013; 2 (ii) 4ax + 3by &#x2013; 3ay &#x2013;4bx (iii) x 3 + y 3 + x 2 y + xy 2

Philoid · Accepted Answer

(i) 7a³ + 7a – 2a² – 2 = 7a(a² + 1) – 2(a² + 1)
= (a² + 1)(7a – 2)

(ii) The terms of 4ax + 3by – 3ay – 4bx can be rearranged and factorized.
4ax – 4bx – 3ay + 3by = 4x(a – b) - 3y(a - b)
= (a – b)(4x – 3y)

(iii) x³ + y³ + x²y + xy²= x³ + x²y + xy² + y³
= x²(x + y) + y² (x + y)
= (x + y) (x² + y²)

Factorise(i) 7a3 + 7a – 2a2 – 2(ii) 4ax + 3by – 3ay –4bx(iii) x3 + y3 + x2y + xy2

Factorise
(i) 7a³ + 7a – 2a² – 2
(ii) 4ax + 3by – 3ay –4bx
(iii) x³ + y³ + x²y + xy²