# Thread: Trying to solve a polynomial

1. Hi,

I'm trying to help a high-school sophomore with a math problem, and unfortunately my algebra days are long behind me. Here's the equation:

x^3-9x-440=0

I know x=8, but I don't know how to find it. I'd appreciate some guidance.

Thanks.

2. Originally Posted by blinky
Hi,

I'm trying to help a high-school sophomore with a math problem, and unfortunately my algebra days are long behind me. Here's the equation:

x^3-9x-440=0

I know x=8, but I don't know how to find it. I'd appreciate some guidance.

Thanks.
use the rational root theorem. if x an integer is a zero it is factor of 440

so look at them 1,2,4,5,8,10,11,20,22,40,44,55,88,110,220,440 ( both plus and minus)

as the sign changes once so there is a positive root
x^3 = 440 + 9x or x^3 > 440 or x > 7
we need to check for values > 7( from the above set)
try 8 and it succeed
so we need not try further
you can devide by x-8 and get a quadratic and solve the same

Originally Posted by kaliprasad
use the rational root theorem. if x an integer is a zero it is factor of 440

so look at them 1,2,4,5,8,10,11,20,22,40,44,55,88,110,220,440 ( both plus and minus)

as the sign changes once so there is a positive root
x^3 = 440 + 9x or x^3 > 440 or x > 7
we need to check for values > 7( from the above set)
try 8 and it succeed
so we need not try further
you can devide by x-8 and get a quadratic and solve the same
Thanks very much. I'll tell her.

Did you know just from looking at the equation that it had to be solved that way? If so, how did you know?

4. Originally Posted by blinky
Thanks very much. I'll tell her.

Did you know just from looking at the equation that it had to be solved that way? If so, how did you know?
factoring is the way out. but if you cannot factor obviously, then we can use the rational root theorem to find the factor. This is one of the ways

Originally Posted by kaliprasad
factoring is the way out. but if you cannot factor obviously, then we can use the rational root theorem to find the factor. This is one of the ways
Thanks again.

6. Originally Posted by blinky
Hi,

I'm trying to help a high-school sophomore with a math problem, and unfortunately my algebra days are long behind me. Here's the equation:

x^3-9x-440=0

I know x=8, but I don't know how to find it. I'd appreciate some guidance.

Thanks.
Knowing that $x=8$, we could them proceed to rewrite the equation as:

$\displaystyle x^3-8x^2+8x^2-64x+55x-440=0$

Factor:

$\displaystyle x^2(x-8)+8x(x-8)+55(x-8)=0$

$\displaystyle (x-8)\left(x^2+8x+55\right)=0$

To get the other 2 roots, we apply the quadratic formula to the quadratic factor:

$\displaystyle x=\frac{-8\pm\sqrt{8^2-4(1)(55)}}{2(1)}=-4\pm\sqrt{39}i$

7. $x^3 - 9x -440$
$=x^3 - 8x^2 + 8x^2 - 64x + 55x - 440$
$=x^2(x - 8) + 8x(x - 8) + 55(x - 8)$
$=(x - 8)(x^2 + 8x + 55)$
$=(x - 8)(x^2 + 8x + 16 -16 + 55)$
$=(x - 8)((x + 4)^2 + 39)$
$=(x - 8)(x + 4)^2 - (-39))$
$=(x - 8)(x + 4)^2 - (\sqrt{39}i)^2$
$=(x - 8)(x + 4 - \sqrt{39}i)(x + 4 + \sqrt{39}i)$

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