3.04 Solving polynomial equations
Solving by factoring fully
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.
Worked example
Question 1
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.
Practice questions
Question 2
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.
Solving using substitutions
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.
Worked example
Question 3
| $p^2+3p-10$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 |
Question 4
| $\left(2x+1\right)^2+2\left(2x+1\right)-3$(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 |
Practice questions
Question 5
Solve for $x$x: $x^4-20x^2+64=0$x4−20x2+64=0 .
Let $p$p be equal to $x^2$x2.
Question 6
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.