Question

Factorise using factor theorem. 2x^3 – 5x^2 – 19x - 42​

Answer 1

➡We can factor the 2x2 - 11x + 14 by using grouping:

Looking at the coefficients of the x2 term and the constant

we have 2 & 14... 2(14) = 28. Look for factors of 28 that

add to the coefficient of the x term ... -11

(-4)(-7) = 28 and -4-7 = -11.

Replace -11x with -4x - 7x and factor by grouping

2x2 - 11x + 14 = 2x2 - 4x - 7x + 14

= 2x(x - 2) -7(x - 2) = (2x-7)(x-2)

answer:

⏩2x3-5x2-19x+42 = (x+3)(2x-7)(x-2)

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.