Posted by Nikki Clark on Wednesday, November 21, 2012 at 1:42pm.
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, Wednesday, November 21, 2012 at 1:59pm
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, Wednesday, November 21, 2012 at 2:06pm
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, Wednesday, November 21, 2012 at 2:59pm
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, Wednesday, November 21, 2012 at 3:13pm
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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