A problem again:::::::MAths...FActor theormes...?

Question

For what values of p does the polybomial (2p+1)x^2 + px + 2p^2 has a factor
(a) x -1, (b) x + 2?
State the value of p for which the polynomial is exactly divisible by x+2 but not by x - 1.


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Answer 1

Given

f(x) = (2p+1)x^2 + px + 2p^2

and f(x) is exactly divisible by x + 2 but not by x - 1

f(x)/(x+2) = (2p+1)x - (3p+2)

In order for the remainder to equal zero we must have:

2p^2 + 6p + 4 = 0
p^2 + 3p + 2 = 0
(p + 2)(p + 1) = 0
p = -2, -1

For p = -2
f(x) = -3x^2 - 2x + 8
This is not exactly divisible by x - 1

For p = -1
f(x) = -x^2 - x + 2
This is exactly divisible by x - 1
So p = -1 is excluded.

The answer is p = -2.

Answer 2

Let q(x) = (2p + 1)x^2 + px + 2p^2

(a) In order for q(x) to have a factor of (x - 1), it must mean
q(1) = 0. However, by the definition of q(x),
q(1) = (2p + 1)(1)^2 + p(1) + 2p^2
q(1) = 2p + 1 + p + 2p^2
q(1) = 2p^2 + 3p + 1, which we can now equate to 0 because
q(1) is equal to 0.

2p^2 + 3p + 1 = 0. This factors into
(2p + 1)(p + 1) = 0. Therefore
p = {-1/2 , -1}

(b) In order for (x + 2) to be a factor of q(x),
q(-2) = 0. So
q(-2) = (2p + 1)[-2]^2 + p(-2) + 2p^2
q(-2) = (2p + 1)4 - 2p + 2p^2
q(-2) = 8p + 4 - 2p + 2p^2
q(-2) = 2p^2 + 6p + 4

2p^2 + 6p + 4 = 0. Divide the equate by 2,
p^2 + 3p + 2 = 0. Therefore,
(p + 2) (p + 1) = 0, so
p = {-2, -1}

Answer 3

the answer is 0, since 1948 thats a clasic problem