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 -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)
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.
Sure! I can help you with that question.
To find the constant value of p, we know that if (x+4) is a factor of 3x^3 + x^2 + px + 24, then substituting x = -4 into the expression should give us 0.
Let's substitute x = -4 into the expression 3x^3 + x^2 + px + 24:
3(-4)^3 + (-4)^2 + p(-4) + 24 = 0
Simplifying this, we get:
-192 + 16 - 4p + 24 = 0
-152 - 4p = 0
To solve for p, we can rearrange this equation as follows:
-4p = 152
p = -152/-4
p = 38
So, the constant value of p is 38.
Now, let's factorize and solve the equation 3x^3 + x^2 + 38x + 24 = 0.
Since we know that (x+4) is a factor, we can use long division or synthetic division to divide the cubic polynomial by (x+4):
--------------------
x+4 | 3x^3 + x^2 + 38x + 24
When we divide it, we find that it is completely divisible without any remainder, which means (x+4) is indeed a factor. Therefore, the polynomial can be factorized as follows:
(x+4)(3x^2 - 11x + 6) = 0
To solve this equation, we set each factor equal to zero:
x + 4 = 0 or 3x^2 - 11x + 6 = 0
From x + 4 = 0, we find that x = -4.
To solve 3x^2 - 11x + 6 = 0, we can either factorize it further or use the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the quadratic equation 3x^2 - 11x + 6 = 0, we have a = 3, b = -11, and c = 6.
Plugging these values into the formula, we get:
x = (-(-11) ± √((-11)^2 - 4 * 3 * 6)) / (2 * 3)
x = (11 ± √(121 - 72)) / 6
x = (11 ± √49) / 6
x = (11 ± 7) / 6
So, we have two possible solutions:
x = 18/6 = 3/1 = 3
x = 4/6 = 2/3
Therefore, the solutions to the equation 3x^3 + x^2 + 38x + 24 = 0 are x = -4, x = 3, and x = 2/3.
I hope this helps! Let me know if you have any further questions.