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.
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.
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
$\displaystyle \chi$ $\displaystyle \sigma$
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.