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# SQA Higher Maths Factorising techniques

3. Factor Theorem

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.

# 1

Identify a factor of f(x) = x⁵ + 1.

# 2

Factorise x² + 5x + 6.

# 3

Solve 22x + x² = –21.

# 4

Identify a factor of f(x)= x³ + x² + x – 3.

# 5

Factorise x² + x – 2.

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