Factorize x^2 + 4x -5 and hence find the roots of the quadratic equation y = x^2 + 4x - 5.

First, identify the integer factors of the constant. In this case -5, has factors -1,5 and 1,-5. Now check to see if the sum of either pair is equal to the value of the coefficient of x. In this case, the value of the coefficient is 4 and the sum of each pair is 4 and -4 respectively, which means the correct pair of factors to use is -1 and 5. This means that the factorized form of x^2 + 4x - 5 = (x - 1)(x + 5). The roots of the equation y = x^2 + 4x - 5 occur when y = 0, which implies we are solving the equation x^2 + 4x - 5 = 0. To do this we factorize the quadratic as so (x - 1)(x + 5) = 0. From this, we know that for this to equal zero either factor could equal zero. This implies that x -1 = 0 and x + 5 = 0. These can easily be solved, the first by adding one to both sides and the second by subtracting five to give an answer of x = 1 and x = -5.

AW

Related Maths GCSE answers

All answers ▸

Factorise 2e – 4f + ex -2fx


How to find the longest side of a right-angled triangle if we are given the two other sides?


What is the nth-term of this sequence? 29, 26, 23, 20, 17...


If a rectangle has sides (x+3) and (2x-1) and an area of 15, find x and the length of each side.