Solving quadratic equations by factorising where rearrangement is required | Maths

Starter quiz

Factorise the expression $x^2 + 2x - 15$

$(x - 3)(x + 5)$ ✓

$(x + 3)(x - 5)$

$(x + 15)(x - 1)$
Which of these is a solution for: $(x - 4)(x + 2) = 0$

$x = 0$

$x = 4$ ✓

$x = -4$

$x = 2$
Which of these is a solution for: $(x + 2)(x - 5) = 0$

$x = 2$

$x = -5$

$x = 5$ ✓

$x = 0$
Which of these is a solution for: $(x + 9)(x + 3) = 0$

$x = 9$

$x = 3$

$x = -3$ ✓

$x = -9$ ✓
Which of these is a solution for: $(x - 6)^2 = 0$

$x = -6$

$x = 6$ ✓

$x = 0$
Factorise the expression $7x^2 + 42x - 49$

$7(x + 1)(x - 7)$

$7(x - 1)(x - 7)$

$7(x - 1)(x + 7)$ ✓

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In order for factorising to be a valid method, the quadratic must equal zero.
To achieve a product of zero, one of the binomial expressions evaluate to zero.
Either expression could be zero so you must find the solution for each expression.
If the quadratic were not equal to zero, you would have uncertainty about what each expression must equal.

Pupils may try to factorise and solve without rearranging first.

Being able to put the factors equal to zero only makes sense if we are using the property that if two values multiply to zero, one of them is zero.

Factorise - To factorise is to express a term as the product of its factors.
Solution (equality) - A solution to an equality with one variable is a value for the variable which, when substituted, maintains the equality between the expressions.