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
- 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
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 -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 3:13pm
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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