FA4, HA4 - Factorising quadratic expressions

FA4, HA4 - Factorising quadratic expressions

Learn to turn a quadratic expression into a product of brackets by reversing expansion. This develops the monic quadratic and difference-of-squares factorisation in FA4 and HA4, including choosing signs and taking a common factor first.

Monic quadratics

Which two brackets expand to x2+7x+12x^2+7x+12? The constant 1212 gives several possible factor pairs. The middle coefficient 77 tells us which pair to choose.

You already know that (x+p)(x+q)=x2+qx+px+pq(x+p)(x+q)=x^2+qx+px+pq. Collecting the middle terms gives x2+(p+q)x+pqx^2+(p+q)x+pq. So the two bracket numbers must have both the correct product and the correct sum.

Monic quadratic

A quadratic expression has highest power 22 in its variable. A monic quadratic has coefficient 11 on its square term, as in x2+7x+12x^2+7x+12. To factorise means to write an expression as a product.

For x2+bx+cx^2+bx+c, seek numbers pp and qq for which pq=cpq=c and p+q=bp+q=b, then write (x+p)(x+q)(x+p)(x+q). The order of the brackets does not matter.

Worked example 1: using both conditions

Question. Factorise x2+7x+12x^2+7x+12.

Plan. List factor pairs of 1212 and find the pair whose sum is 77.

Pair with product 12Sum1,12132,683,47\begin{array}{c|c} \text{Pair with product }12&\text{Sum}\\ 1,12&13\\ 2,6&8\\ 3,4&7 \end{array}

The numbers 33 and 44 satisfy both conditions.

Answer. (x+3)(x+4)(x+3)(x+4).

Check. Expanding gives x2+4x+3x+12=x2+7x+12x^2+4x+3x+12=x^2+7x+12, exactly the original expression.

Keep the constant 1212, but change the middle coefficient from 77 to 88. The factor pair must change.

Worked example 2: the same product, a different sum

Question. Factorise x2+8x+12x^2+8x+12.

Plan. Reuse the product pairs, selecting by the new sum.

2×6=12,2+6=82\times6=12,\qquad 2+6=8

Therefore choose p=2p=2 and q=6q=6.

Answer. (x+2)(x+6)(x+2)(x+6).

Check. (x+2)(x+6)=x2+6x+2x+12=x2+8x+12(x+2)(x+6)=x^2+6x+2x+12=x^2+8x+12. As a numerical check, at x=1x=1 both forms give 2121.

Getting only the constant right is not enough.

Watch out

(x+1)(x+12)(x+1)(x+12) has constant 1212, but its middle coefficient is 1313. Check the sum as well as the product before choosing brackets.

Use the same search on a new constant.

Signed factors

The sum-and-product rule also works with negative numbers; the constant tells you whether the two signs agree.

Constant ccSigns of pp and qqHow the sum helps
PositiveBoth positive or both negativePositive bb: both positive. Negative bb: both negative.
NegativeOne positive and one negativeThe sign of bb belongs to the number with greater magnitude.

For example, −3-3 and −4-4 have product +12+12 and sum −7-7. Therefore x2−7x+12=(x−3)(x−4)x^2-7x+12=(x-3)(x-4). Expanding produces x2−4x−3x+12x^2-4x-3x+12, so both conditions are satisfied.

When the constant is negative, choose the magnitudes first, then assign signs to obtain the required sum.

Worked example 3: opposite signs and a positive sum

Question. Factorise x2+x−20x^2+x-20.

Plan. Find factors of 2020 whose difference is 11, then make the larger one positive.

5×(−4)=−20required product5+(−4)=1required sum\begin{aligned} 5\times(-4)&=-20 &&\text{required product}\\ 5+(-4)&=1 &&\text{required sum} \end{aligned}

Answer. (x+5)(x−4)(x+5)(x-4).

Check. (x+5)(x−4)=x2−4x+5x−20=x2+x−20(x+5)(x-4)=x^2-4x+5x-20=x^2+x-20.

Now change only the sign of the middle term. The magnitudes stay 44 and 55, but their signs switch.

Continue this lesson.

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