Question

Factorise using factor theorem:x² + 4x-5 (iii) y2-2y-15​

Answer 1

Step-by-step explanation:

(1)

${x}^{2} + 4x - 5 \\ {x}^{2} + 5x - x - 5 \\ x(x + 5) \: - 1(x + 5) \\ (x + 5) \: (x - 1) \\ \\ x = - 5 \: \: \: \: \: \: \: \: \: x = 1$

(2)

${y}^{2} - 2y - 15 \\ {y}^{2} - 5y + 3y - 15 \\ y(y - 5) \: + 3(y - 5) \\ (y - 5) \: \: (y + 3) \\ \\ y = 5 \: \: \: \: \: \: \: \: y = - 3$

Answer 2

$1) \: \: \: {x}^{2} + 4x - 5$

$= {x}^{2} + (5 - 1)x - 5$

$= {x}^{2} + 5x - x - 5$

$= x(x + 5) - 1(x + 5)$

$= (x - 1)(x + 5)$

$2) \: \: \: {y}^{2} - 2y - 15$

$= {y}^{2} - (5 - 3)y - 15$

$= {y}^{2} - 5y + 3y - 15$

$= y(y - 5) + 3(y - 5)$

$= (y - 5)(y + 3)$