Asked by Nikki Clark

Hey guys, need some help with a question ive been stuck on for a looooooong time!

The question is:
Given that (x+4) is a factor of the expression 3x^3+x^2+px+24, find the constant value of p. Hence factorise and solve the equation 3x^3+x^2+px+24=0

Any ideas??????

Answers

Answered by Count Iblis
If x+4 is a factor of a polynomial
q(x), then you have:

q(x) = (x+4)r(x)

for some polynomial r(x). If you put
x = -4, you get:

q(-4) = 0


Putting x = -4 in

3x^3+x^2+px+24

and equating to zero gives:

-3 2^6 + 2^4 - 4 p + 3*8 = 0 ----->

p = -3 2^4 + 4 + 3*2 = -38
Answered by Nikki Clark
im really lost and really want to understand this question. Is there any other way of explaining it. Thank you for your time
Answered by Steve
you could do a synthetic division. The values along the bottom row are

3 -11 p+44 -4p-152

that means that

3x^3+x^2+px+24 = (x+4)(3x^2-11x+p+44) + (-4p-152)

the remainder is -4p-152, which we want to be zero if (x+4) is a factor.

so, p = -38, and our cubic is

3x^3 + x^2 - 38x + 24 = (x+4)(3x^2-11x+6)
Answered by Damon
If (x+4) is a factor then if you use -4 for x the original polynomial is zero.

Just plug in -4 for x like they did.

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