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# SQA National 5 Maths Factorising techniques 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 (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. # ✅

Suppose we have 5 + x = 1 + y. Rearrange the equation so that y is the subject. # ✅

Solve s – 4 = b + 3 for b when s = 4. # ✅

Factorise x² + 5x + 6. # ✅

Solve x² + 5x + 4 = 0. # ✅

Solve x² + 5x = 6. # ✅

Solve x² + 6x + 9 = 0 using the quadratic formula. # ✅

Solve 4x² + 3x – 7 = 0 using the quadratic formula. # ✅

Solve 22x + x² = –21.  Have you found the questions useful?