Find the value of p for which x³+4x²-px+8 is exactly divisible by (x-2) is.

Given Polynomial P(x) = x^3+4x^2-px+8

Given that P(x) is exactly divisible by (x-2) then by

factor theorem it is a factor and P(2) = 0

⇛ (2)^3+4(2)^2-p(2)+8 = 0

⇛8+4(4)-2p+8 = 0

⇛ 8+16-2p+8 = 0

⇛ 32-2p = 0

⇛32 = 2p

⇛ 2p = 32

⇛ p = 32/2

⇛ p = 16

Answer:- The value of p is 16.

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Similar Questions

Find the value of p for which the polynomial 3x^3 -x^2 + px +1 is exactly divisible by x-1, hence factorise the polynomial...

16.

Step-by-step explanation:

By the Factor Theorem if a polynomial is divisible by (x - a) then f(a) = 0 so here we have:

2^3 + 4(2)^2 - 2p + 8 = 0

-2p + 32 = 0

p = 16.