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Factors of quadratics

CraigsMaths
Mathematics Teaching for Learning
Maths methods A, Quadratic Equations, Trinomial factors
August 13th, 2008 by Craig Rose
Permalink: http://www.craigsmaths.com/quadratic-equations/factors-of-quadratics/

Maths methods A, Quadratic Equations, Trinomial factors
August 13th, 2008 by Craig Rose
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When factorising quadratics (or any trinomial) first identify the form of the expression and then apply the most direct method. If the expression is not in a recognisable form then manipulate it so that it is.

Common factors of coefficients

Before proceeding first extract any common factors of the coefficients. Then factorise the remaining unfactored expression.

Perfect squares

For a perfect square the factors can be determined directly:

$a^{2}x^{2}+2abx+b^{2} = \left(ax+b\right)^{2}$

Difference of squares

Where the expression can be expressed as a difference of two squares the factors are also determined directly:

$ax^{2}-b = \left(\sqrt{a}x-\sqrt{b}\right) \left(\sqrt{a}x+\sqrt{b}\right)$

General form with $a=1$

$x^{2}+bx+c$

Find two factors, $n$ and $m$ , of the constant part $c$ that add to equal $b$ . Watch the signs of the coefficient.

$x^{2}+bx+c=\left(x+n\right)\left(x+m\right)$

Where $nm=c$ and $n+m=b$ .

Completing the square

Example

Factorise $2x^{2}-32x+39$

Step 1

Divide the coefficients of the $x^{2}$ and the $x$ terms by a number that will leave the $x^{2}$ term with a coefficient of $1$ .

In this case we divide by 2.

$2\left(x^{2}-16x\right)+39$

Step 2

Add the square of half the coefficient of the $x$ term to the grouped $x$ and $x^{2}$ terms. Subtract this amount times the constant in front of the brackets from the constant term.

In this case we want to add:

$\left(\frac{16}{2}\right)^{2}=64$