Asked by Nikki Clark
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??????
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??????
Answers
Answered by
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
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
Answered by
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
Answered by
Steve
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)
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)
Answered by
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.
Just plug in -4 for x like they did.