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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 $\left(4x^2-9\right)\left(x^2+5x-6\right)=0$(4x29)(x2+5x6)=0.

Think: The two quadratic factors can both be factored further. Once it is fully factored, we can simply apply the zero product property.


$\left(4x^2-9\right)\left(x^2+5x-6\right)$(4x29)(x2+5x6) $=$= $0$0 State the original equation
$\left(2x-3\right)\left(2x+3\right)\left(x+6\right)\left(x-1\right)$(2x3)(2x+3)(x+6)(x1) $=$= $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$2x3=0 $2x+3=0$2x+3=0 $x+6=0$x+6=0 $x-1=0$x1=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:


  1. 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
Solve the equation $x^4+3x^2-10=0$x4+3x210=0.
Think: If we look at replacing every $x^2$x2 in the equation with a $p$p then we can rewrite the equation as a more familiar quadratic where $p$p is the variable. 
Notice that $x^4=\left(x^2\right)^2$x4=(x2)2, so this will become $p^2$p2. We can then substitute $3x^2$3x2 with $3p$3p
Do: Our full substitution and letting $p=x^2$p=x2 gives $x^4+3x^2-10=p^2+3p-10$x4+3x210=p2+3p10. From here we could solve in any number of ways! Let's solve for $p$p by completing the square. 
$p^2+3p-10$p2+3p10 $=$= $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 $=$= $\pm\frac{7}{2}-\frac{3}{2}$±7232
$p$p $=$= $\frac{7}{2}-\frac{3}{2}$7232  and $\frac{-7}{2}-\frac{3}{2}$7232
$p$p $=$= $\frac{4}{2}$42  and $\frac{-10}{2}$102
$p$p $=$= $2$2   and  $-5$5
BUT - remember that we made a substitution, and $p=x^2$p=x2. So we haven't finished yet, we still need to solve for $x$x. This is one of the most common mistakes, not finishing the question. 
$p$p $=$= $2$2
$x^2$x2 $=$= $2$2
$x$x $=$= $\pm\sqrt{2}$±2
$x^2$x2 $=$= $-5$5
Since there is no real number that gives $-5$5 when squared, $x^2=-5$x2=5 has no real solutions. So the real roots to this function are $x=\sqrt{2}$x=2 or $x=-\sqrt{2}$x=2.
Question 4
What are the solutions of the quadratic equation $\left(2x+1\right)^2+2\left(2x+1\right)-3=0$(2x+1)2+2(2x+1)3=0
Think: We could distribute it completely, combine like terms and then solve. This would involve a lot of extra algebraic manipulation. Or, we could make a clever substitution...
Do: Let's see what happens when we let $j=2x+1$j=2x+1.
$\left(2x+1\right)^2+2\left(2x+1\right)-3$(2x+1)2+2(2x+1)3 $=$= $0$0 substitute $j=2x+1$j=2x+1
$j^2+2j-3$j2+2j3 $=$= $0$0  
$\left(j+3\right)\left(j-1\right)$(j+3)(j1) $=$= $0$0  
$j+3$j+3 $=$= $0$0 Where $j=-3$j=3
$j-1$j1 $=$= $0$0 Where $j=1$j=1
Remember that $j$j $=$= $2x+1$2x+1  
$j$j $=$= $-3$3 becomes
$2x+1$2x+1 $=$= $-3$3  
$2x$2x $=$= $-4$4  
$x$x $=$= $-2$2  
$j$j $=$= $1$1 becomes
$2x+1$2x+1 $=$= $1$1  
$2x$2x $=$= $0$0  
$x$x $=$= $0$0  
So the roots of the quadratic equation $\left(2x+1\right)^2+2\left(2x+1\right)-3=0$(2x+1)2+2(2x+1)3=0 are $x=-2$x=2 or $x=0$x=0.


Practice questions

Question 5

Solve for $x$x: $x^4-20x^2+64=0$x420x2+64=0 .

Let $p$p be equal to $x^2$x2.

Question 6

Solve the following equation for $x$x:


You may let $p=9x+10$p=9x+10.


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