Question

Find p if x^3-3x^2+px-8 is divisible by (x+2)

Answer 1

★Given: -

p(x) = x³ -3x² +px - 8 is divisible by (x+2)

★To find:-

value of p

★Solution:-

We know that,

x+2 = 0, x = -2

x+2 is divisible by p(x) , then p(-2) = 0 by factor theorem.

p(-2) = (-2)³- 3(-2)² + (-2)p -8

= -8-12-2p-8

= -28 -2p

-28 -2p = 0

-2p = 28

$p = \frac{28}{-2}\\ \\ \\ {\pink{\boxed{p = -14}}}$

★Additional Information -

$\purple{Remainder \:theorem:}$ Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).

Answer 2

$x^{2}$Answer:

Step-by-step explanation:

Given the polynomial f(x)=x^{3}+4x^{2}-px+8

For f(x) to be exactly divisible by 2, we must have f(2)=0 by remainder and factor theorem

Since given that, (x – 2) is exactly divisible by 2.

Thus, x = 2

Substituting x=2 in the given expression, we get,

f(2)=2^{3}+4*2^{2} -2p+8=0

8+16-2p+8=0

16+16-2p=0

32-2p=0

p-16

Which gives p=16  

Thus, the value of p will be 16.

Hope it helps uh...

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