# Question Video: Factorising by Grouping Mathematics • 9th Grade

Factorise fully 𝑥³ + 𝑥 − 130.

04:32

### Video Transcript

Factorise fully 𝑥 cubed plus 𝑥 minus 130.

If we let 𝑓 of 𝑥 equal 𝑥 cubed plus 𝑥 minus 130, we can find one factor using the factor theorem. The factor theorem states that if 𝑓 of 𝑎 is equal to zero, then 𝑥 minus 𝑎 is a factor of 𝑓 of 𝑥. We need to find a number for 𝑥 that we can substitute into the equation to give us an answer of zero.

We could do this using a trial and improvement method: firstly, substituting in 𝑥 equals one then 𝑥 equals two and going through the positive integers and then the negative integers. However, this would take us a long time. It is often worth doing some calculations in your head first.

In this case, 𝑓 of five is equal to five cubed plus five minus 130. Five cubed is equal to 125. Therefore, we are left with 125 plus five minus 130. As this is equal to zero, we can say that 𝑥 minus five is a factor. As 𝑥 minus five is a factor, we need to divide 𝑓 of 𝑥 𝑥 cubed plus 𝑥 minus 130 by 𝑥 minus five.

There are lots of methods of doing this. But we’ll use a long division method, where 𝑥 minus five is the divisor, 𝑥 cubed plus 𝑥 minus 130 is the dividend, and the answer will be the quotient. You will notice that we have added zero 𝑥 squared into the dividend as there was no 𝑥 squared term in the function 𝑓 of 𝑥.

Our first step is to divide the first term of the dividend 𝑥 cubed by 𝑥, the first term of the divisor. 𝑥 cubed divided by 𝑥 is equal to 𝑥 squared. We now multiply 𝑥 squared by 𝑥 minus five and subtract. 𝑥 squared multiplied by 𝑥 is 𝑥 cubed and 𝑥 squared multiplied by negative five is negative five 𝑥 squared. Subtracting these two lines leaves us with five 𝑥 squared.

Our next step is to bring down the next term — in this case plus 𝑥. We now need to repeat the process. Dividing five 𝑥 squared by 𝑥 gives us five 𝑥. We then need to multiply five 𝑥 by 𝑥 minus five. This gives us five 𝑥 squared minus 25𝑥. And finally, subtracting these two lines gives us 26𝑥 as 𝑥 minus negative 25𝑥 is 26𝑥.

We then drop down the final term — in this case negative 130. Dividing 26𝑥 by 𝑥 gives us 26 and multiplying 26 by 𝑥 minus five is equal to 26𝑥 minus 130. When we subtract these two lines, we get an answer of zero as 𝑥 minus five was a factor of 𝑓 of 𝑥. There will be no remainder.

This means that we can factorise 𝑥 cubed plus 𝑥 minus 130 in the form 𝑥 minus five multiplied by 𝑥 squared plus five 𝑥 plus 26. We would normally then factorise the quadratic into two more brackets or parentheses. However, the quadratic 𝑥 squared plus five 𝑥 plus 26 cannot be factorised.

So our final answer is 𝑥 minus five multiplied by 𝑥 squared plus five 𝑥 plus 26.