Question

Factorise 2x^3 - 5x^2 + 19x - 42

Answer 1

ANSWER:

$2 {x}^{3} - 5 {x}^{2} - 19x + 42$

$2 {x}^{3} - 4 {x}^{2} - {x}^{2} + 2x - 21x + 42$

$2 {x}^{2} (x - 2) - x (x - 2) - 21(x - 2)$

$(x - 2)(2 {x}^{2} - x - 21)$

$(x - 2)(2 {x}^{2} + 6 x - 7x- 21)$

$(x - 2){(2x(x + 3) - 7(x + 3) )}$

$(x - 2)(2x - 7)(x + 3)$

Thanks!

Answer 2

Answer:

Answer:

factors of 42 ----> ±1, ±2, ±3, ±6, ±7

$f(1) = 2(1) {}^{3} - 5(1) {}^{2} - 19(1) + 42 \\ \\ = 2 - 5 - 19 + 42 ≠ 0$

$f( - 1) = 2( - 1) {}^{3} - 5( -1 ) - 19( - ) + 42 \\ \\ = - 3 - 5 + 19 + 42 ≠ 0$

$f( - 2) = 2( - 2) {}^{3} - 5( - 2) {}^{2} - 19( - 12) + 42 \\ \\ = - 16 - 20 + 19 + 42 \\ \\ ≠ 0$

$f(2) = 2(2) {}^{3} - 5(2) {}^{2} - 19(2) + 42 \\ \\ = 16 - 20 - 38 + 4$

x = 2 is zero

Therefore, (x – 2) is factor.