As we mentioned when we were looking at sketching polynomials, we can still use the zero product property to solve polynomial equations. We just need the polynomial to be fully factored. This might be done for us, or we might have to do some factoring.
Solve $\left(4x^2-9\right)\left(x^2+5x-6\right)=0$(4x2−9)(x2+5x−6)=0.
Think: The two quadratic factors can both be factored further. Once it is fully factored, we can simply apply the zero product property.
Do:
$\left(4x^2-9\right)\left(x^2+5x-6\right)$(4x2−9)(x2+5x−6) | $=$= | $0$0 | State the original equation |
$\left(2x-3\right)\left(2x+3\right)\left(x+6\right)\left(x-1\right)$(2x−3)(2x+3)(x+6)(x−1) | $=$= | $0$0 | Factor fully from a difference of squares and a simple trinomial |
We can now apply to zero product property to get:
$2x-3=0$2x−3=0 | $2x+3=0$2x+3=0 | $x+6=0$x+6=0 | $x-1=0$x−1=0 |
$2x=3$2x=3 | $2x=-3$2x=−3 | $x=-6$x=−6 | $x=1$x=1 |
$x=\frac{3}{2}$x=32 | $x=\frac{-3}{2}$x=−32 |
The solutions are $x=\frac{3}{2}$x=32, $x=\frac{-3}{2}$x=−32, $x=-6$x=−6 and$x=1$x=1.
Solve the following equation:
$\left(x^2-16\right)\left(x^2+12x+36\right)=0$(x2−16)(x2+12x+36)=0
Write all solutions on the same line, separated by commas.
Sometimes functions don't even look like quadratics, but with some clever substitutions, we can make it look like a quadratic to enable us to solve them.
$p^2+3p-10$p2+3p−10 | $=$= | $0$0 |
$p^2+3p$p2+3p | $=$= | $10$10 |
$p^2+3p+\left(\frac{3}{2}\right)^2$p2+3p+(32)2 | $=$= | $10+\left(\frac{3}{2}\right)^2$10+(32)2 |
$\left(p+\frac{3}{2}\right)^2$(p+32)2 | $=$= | $10+\frac{9}{4}$10+94 |
$\left(p+\frac{3}{2}\right)^2$(p+32)2 | $=$= | $\frac{49}{4}$494 |
$p+\frac{3}{2}$p+32 | $=$= | $\pm\frac{7}{2}$±72 |
$p$p | $=$= | $\pm\frac{7}{2}-\frac{3}{2}$±72−32 |
$p$p | $=$= | $\frac{7}{2}-\frac{3}{2}$72−32 and $\frac{-7}{2}-\frac{3}{2}$−72−32 |
$p$p | $=$= | $\frac{4}{2}$42 and $\frac{-10}{2}$−102 |
$p$p | $=$= | $2$2 and $-5$−5 |
$p$p | $=$= | $2$2 |
Then | ||
$x^2$x2 | $=$= | $2$2 |
$x$x | $=$= | $\pm\sqrt{2}$±√2 |
AND | ||
$x^2$x2 | $=$= | $-5$−5 |
$\left(2x+1\right)^2+2\left(2x+1\right)-3$(2x+1)2+2(2x+1)−3 | $=$= | $0$0 | substitute $j=2x+1$j=2x+1 |
$j^2+2j-3$j2+2j−3 | $=$= | $0$0 | |
$\left(j+3\right)\left(j-1\right)$(j+3)(j−1) | $=$= | $0$0 | |
So | |||
$j+3$j+3 | $=$= | $0$0 | Where $j=-3$j=−3 |
$j-1$j−1 | $=$= | $0$0 | Where $j=1$j=1 |
Remember that $j$j | $=$= | $2x+1$2x+1 | |
Then | |||
$j$j | $=$= | $-3$−3 | becomes |
$2x+1$2x+1 | $=$= | $-3$−3 | |
$2x$2x | $=$= | $-4$−4 | |
$x$x | $=$= | $-2$−2 | |
And | |||
$j$j | $=$= | $1$1 | becomes |
$2x+1$2x+1 | $=$= | $1$1 | |
$2x$2x | $=$= | $0$0 | |
$x$x | $=$= | $0$0 |
Solve for $x$x: $x^4-20x^2+64=0$x4−20x2+64=0 .
Let $p$p be equal to $x^2$x2.
Solve the following equation for $x$x:
$3\left(9x+10\right)^2+19\left(9x+10\right)+20=0$3(9x+10)2+19(9x+10)+20=0
You may let $p=9x+10$p=9x+10.