very complicated maths
posted by Nikki Clark on .
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??????

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 
im really lost and really want to understand this question. Is there any other way of explaining it. Thank you for your time

you could do a synthetic division. The values along the bottom row are
3 11 p+44 4p152
that means that
3x^3+x^2+px+24 = (x+4)(3x^211x+p+44) + (4p152)
the remainder is 4p152, 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^211x+6) 
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.