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.

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- Sep 6th 2010, 09:34 AMMack4545polynomial functions of Higher Degree
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. - Sep 6th 2010, 09:50 AMchisigma
Setting $\displaystyle x=-2$ You obtain $\displaystyle y=0$... that means that $\displaystyle 4\ x^{3} + 4\ x^{2} - 7\ x + 2$ contains $\displaystyle x+2$ as a factor... etc...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Sep 6th 2010, 10:04 AMMack4545
- Sep 6th 2010, 10:27 AMchisigma
If You insert $\displaystyle x=-2$ into the expression $\displaystyle y(x)= 4\ x^{3} + 4\ x^{2} - 7\ x + 2$ You obtain $\displaystyle y(-2) = -32 + 16 +14 +2 =0$ ...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Sep 6th 2010, 11:07 AMHallsofIvy
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 $\displaystyle \pm 1$, $\displaystyle \pm 2$, or $\displaystyle \pm 4$. The constant term is 2 so any rational root must have numerator $\displaystyle \pm 1$ or $\displaystyle \pm 2$. That means that the only possible rational roots are $\displaystyle \pm 1$, $\displaystyle \pm \frac{1}{2}$, $\displaystyle \pm\frac{1}{4}$, $\displaystyle \pm 2$, $\displaystyle \pm\frac{2}{2}= \pm 1$, and $\displaystyle \pm\frac{2}{4}= \pm\frac{1}{2}$. Trying each of those shows that only x= -2 really is a root.