When solving linear equations we generally want to rearrange it, and make some variable the 'subject' of the formula. Something is the subject of the formula if it appears by itsel on one side of the equation. For example, if y = 2x + 3, then y is the subject. To solve an equation for y, we want to isolate it like that on one side, and then evaluate the terms on the other side with numbers we know.

When given a quadratic such as (x + 1)(x + 2), we can multiply out the brackets using FOIL (meaning first, outside, inside, last), giving x² + 2x + x + 2, which is just x² + 3x + 2. However, for solving a quadratic it's more useful to do this process in reverse. Given some quadratic like x² + 3x + 2, we want to find 2 values to fit in (x + _)(x + _). We know these values need to equal 2 when multiplied together, and when summed they need to equal 3. Once we spot the values 1 and 2 for this, we have (x + 1)(x + 2), and it's very easy to solve an equation like (x + 1)(x + 2) = 0, because this just gives our solutions x = –1 and –2.

We have a quadratic equation of the form ax² + bx + c = 0, where a, b, and c are just coefficients. The solutions to the quadratic are given by the quadratic formula: x = (–b +– √(b² – 4ac))/2a.

For a polynomial f(x), if f(a) = 0 then we know that (x – a) is a factor of the polynomial. Likewise, is we know (x – a) is a factor, then f(a) = 0. While not particularly useful for quadratics, this can prove helpful in factorising more difficult polynomials of order 3 or above.

​

Notice that f(–1) = 0, hence by the factor theorem we know (x + 1) is a factor of the polynomial.

​

We want something of the form (x + _)(x + _) and we know when multiplied together, the values in the gaps should equal 6. They could either be 1 and 6, or 2 and 3 (or also –1 and –6, or –2 and –3). However the only pair of values here that add up to 5 is 2 and 3, so we have (x + 2)(x + 3).

​

One needs to rearrange and be aware of the ordering of terms. We have a = 1, b = 22 and c = 21, which in the quadratic formula gives our solutions x = –1 and x = –21.

​

Notice that f(1) = 0, hence by the factor theorem (x – 1) is a factor of the polynomial.

​

Can first rearrange to b = s – 7. Then substitute 4 in for s, which gives the solution b = –3.