Question

Factorise 2x^3 – 5x^2 – 19x + 42​

Answer 1

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.