Question

Find b if 2x^3 + bx²-7x -12 is perfectly dívisble by
x+2​

Answer 1

Answer:

14

Step-by-step explanation:

Here, p(x) = 2x³+bx²-7x-12

g(x) = x+2

As x+2 is a factor of p(x) so,

Put g(x) and p(x) = 0

g(x) = 0

x+2 = 0

So, x= -2 in p(x)

p(x) = 0

that means:-

2x³ + bx² -7x -12 = 0

Put x = -2

2(-2)³ + b(-2)² -7(-2)-12=0

-16 +4b + 14 -12 = 0

-16-12+14+4b= 0

-28+14+b= 0/4

-14+b= 0

b= 0+14

b= 14

Hope it helps....

Thanks....

Answer 2

Answer:

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${2x}^{3} + bx {}^{2} - 7x - 12 \\ \ \\ 2( - 2) {}^{3} + {b( - 2)}^{2} - 7( - 2) - 12 \\ \\ 2( - 8) + b( - 4) - 7( - 2) - 12 \\ \\ 16 - 4b - 14 - 12 \\ \\ 4b - 14 - 4$

$4b - 18$

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