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Factorising techniques for SQA Higher Maths

Factorising techniques

This page covers the following topics:

1. Factorising quadratic equations
2. The quadratic equation
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 xยฒ + 2x + x + 2, which is just xยฒ + 3x + 2. However, for solving a quadratic it's more useful to do this process in reverse. Given some quadratic like xยฒ + 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 = โ€“1 and โ€“2.

Factorising quadratic equations

We have a quadratic equation of the form axยฒ + bx + c = 0, where a, b, and c are just coefficients. The solutions to the quadratic are given by the quadratic formula: x = (โ€“b +โ€“ โˆš(bยฒ โ€“ 4ac))/2a.

The quadratic equation

For a polynomial f(x), if f(a) = 0 then we know that (x โ€“ a) is a factor of the polynomial. Likewise, is we know (x โ€“ 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.

Factor Theorem

1

Identify a factor of f(x) = xโต + 1.

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

Identify a factor of f(x) = xโต + 1.

2

Factorise xยฒ + 5x + 6.

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 โ€“1 and โ€“6, or โ€“2 and โ€“3). However the only pair of values here that add up to 5 is 2 and 3, so we have (x + 2)(x + 3).

Factorise xยฒ + 5x + 6.

3

Solve 22x + xยฒ = โ€“21.

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 = โ€“1 and x = โ€“21.

Solve 22x + xยฒ = โ€“21.

4

Identify a factor of f(x)= xยณ + xยฒ + x โ€“ 3.

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

Identify a factor of f(x)= xยณ + xยฒ + x โ€“ 3.

5

Factorise xยฒ + x โ€“ 2.

One needs to notice that โ€“1 and 2 multiplied together give โ€“2, but when added together give 1 (the coefficient of x in our equation). The factorised equation is therefore (x โ€“ 1)(x + 2).

Factorise xยฒ + x โ€“ 2.

End of page

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