polynomial functions of Higher Degree

**Mack4545** · Sep 6th 2010, 09:34 AM

Hello I am asked the following: "Find the zeros algebraically".

y=4x^3+4x^2-7x+2

I know with the four terms I am supposed to group, however Grouping does not work with this problem. How do I factor this to find the zeros. Thanks.

**chisigma** · Sep 6th 2010, 09:50 AM

Setting $x=-2$ You obtain $y=0$ ... that means that $4\ x^{3} + 4\ x^{2} - 7\ x + 2$ contains $x+2$ as a factor... etc...

Kind regards

$\chi$ $\sigma$

**Mack4545** · Sep 6th 2010, 10:04 AM

Originally Posted by chisigma

Setting $x=-2$ You obtain $y=0$ ... that means that $4\ x^{3} + 4\ x^{2} - 7\ x + 2$ contains $x+2$ as a factor... etc...

Kind regards

$\chi$ $\sigma$

Thank you very much for your reply, however I am wondering how exactly did you find that one of the zeros is -2?

Thanks

**chisigma** · Sep 6th 2010, 10:27 AM

Originally Posted by Mack4545

Thank you very much for your reply, however I am wondering how exactly did you find that one of the zeros is -2?

Thanks

If You insert $x=-2$ into the expression $y(x)= 4\ x^{3} + 4\ x^{2} - 7\ x + 2$ You obtain $y(-2) = -32 + 16 +14 +2 =0$ ...

Now You devide $y(x)= 4\ x^{3} + 4\ x^{2} - 7\ x + 2$ by $x+2$ and obtain a second degree polynomial whose zeroes can be found in a weel known way...

Kind regards

$\chi$ $\sigma$

**HallsofIvy** · Sep 6th 2010, 11:07 AM

Very useful here is the "rational roots theorem": If m/n is a rational root of a polynomial equation with integer coefficients, then the numerator, m, must evenly divide the constant term and the denominator, n, must evenly divide the leading coefficient.

Of course, that doesn't guarentee that there are any rational roots but it is always worth trying.

Here the leading coefficient is 4 so any rational root must have denominator $\pm 1$ , $\pm 2$ , or $\pm 4$ . The constant term is 2 so any rational root must have numerator $\pm 1$ or $\pm 2$ . That means that the only possible rational roots are $\pm 1$ , $\pm \frac{1}{2}$ , $\pm\frac{1}{4}$ , $\pm 2$ , $\pm\frac{2}{2}= \pm 1$ , and $\pm\frac{2}{4}= \pm\frac{1}{2}$ . Trying each of those shows that only x= -2 really is a root.

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