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??????

very complicated maths 
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 
very complicated maths 
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

very complicated maths 
Steve,
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) 
very complicated maths 
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.