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September 3, 2014

September 3, 2014

Posted by **Nikki Clark** on Wednesday, November 21, 2012 at 1:42pm.

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??????

- very complicated maths -
**Count Iblis**, Wednesday, November 21, 2012 at 1:59pmIf 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

- very complicated maths -
**Nikki Clark**, Wednesday, November 21, 2012 at 2:06pmim really lost and really want to understand this question. Is there any other way of explaining it. Thank you for your time

- very complicated maths -
**Steve**, Wednesday, November 21, 2012 at 2:59pmyou 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)

- very complicated maths -
**Damon**, Wednesday, November 21, 2012 at 3:13pmIf (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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