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Middle Years

2.03 Solve by factorisations

Lesson

Solve quadratic equations using factorisation

  1. Gather together all the terms on one side of the equation so that the other side is $0$0.
  2. Factorise these terms using any appropriate techniques.
  3. Use the null factor law to split the quadratic equation into two linear equations.
  4. Solve the linear equations to get the solutions to the original quadratic equation.

 

Worked example

Solve $2x^2+4x+2=x^2-4x-13$2x2+4x+2=x24x13 for $x$x.

Think: Since there are square terms in the equation this is a quadratic equation, so we can use the method above.

Do:

$2x^2+4x+2$2x2+4x+2 $=$= $x^2-4x-13$x24x13

 

$x^2+8x+15=0$x2+8x+15=0 $=$= $0$0

Gathering all of the terms on the left hand side and simplifying

$\left(x+5\right)\left(x+3\right)$(x+5)(x+3) $=$= $0$0

Factorising the quadratic expression on the left hand side

Note that $5+3=8$5+3=8 and $5\times3=15$5×3=15

Now we use the null factor law to split this quadratic equation into two linear equations.

$x+5$x+5 $=$= $0$0

Using one factor

$x$x $=$= $-5$5

Solving for $x$x

$x+3$x+3 $=$= $0$0

Using the other factor

$x$x $=$= $-3$3

Solving for $x$x

So the solutions are $x=-5$x=5 and $x=-3$x=3.

 

Summary

We can solve many quadratic equations by using this method:

  1. Gather together all the terms on one side of the equation so that the other side is $0$0.
  2. Factorise these terms using any appropriate techniques.
  3. Use the null factor law to split the quadratic equation into two linear equations.
  4. Solve the linear equations to get the solutions to the original quadratic equation.

Practice questions

Question 1

The equation $2y-6y^2=0$2y6y2=0 has two solutions. Solve for both values of $y$y, writing your answers on the same line separated by a comma.

Question 2

Solve $x^2=13x+114$x2=13x+114 for $x$x.

Enter each solution on the same line, separated by a comma.

Question 3

Solve $x^2+13x+42=0$x2+13x+42=0 for $x$x.

Enter each solution on the same line, separated by a comma.

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