E2.2 - Factorising quadratics (Extended)

E2.2 - Factorising quadratics (Extended)

Which two brackets expand to x2+7x+12x^2+7x+12? The middle coefficient and the constant give two clues, and both must fit. This Extended E2.2 lesson develops quadratic factorisation, then combines it with extracting a common factor to factorise cubic expressions fully.

Monic quadratics

A rectangle with sides x+3x+3 and x+4x+4 has area (x+3)(x+4)(x+3)(x+4). Splitting each side shows four areas: x2x^2, 3x3x, 4x4x and 12. The two middle areas combine to give 7x7x.

[DIAGRAM: asset_name: Quadratic area model; asset_slug: quadratic_area_model; caption: The two side additions multiply to 12 and add to 7.; description: A rectangle partitioned into four regions with side segments x and 3 horizontally, and x and 4 vertically. The four regions are labelled x squared, 3x, 4x and 12. Full sides are x plus 3 and x plus 4. The illustration uses positive x; the resulting algebraic identity holds for every real x.]
The two side additions multiply to 12 and add to 7.

Monic quadratic

A quadratic expression has highest power 2 and a non-zero coefficient of that power. It is monic when the coefficient of x2x^2 is 1, as in x2+7x+12x^2+7x+12.

Expanding two brackets explains the clues:

(x+p)(x+q)=x2+(p+q)x+pq.(x+p)(x+q)=x^2+(p+q)x+pq.

To reverse this expansion, find numbers whose product is the constant and whose sum is the coefficient of xx. A positive product needs equal signs; a negative product needs opposite signs. The area picture uses positive lengths, but expansion proves the identity for negative values of xx too.

Worked example 1: two positive constants

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

Plan. Find a factor pair of 12 that adds to 7.

1+12=13wrong sum2+6=8wrong sum3+4=7required pairx2+7x+12=(x+3)(x+4)\begin{aligned} 1+12&=13 &&\text{wrong sum}\\ 2+6&=8 &&\text{wrong sum}\\ 3+4&=7 &&\text{required pair}\\ x^2+7x+12&=(x+3)(x+4) \end{aligned}

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

Check. Expansion gives x2+4x+3x+12=x2+7x+12x^2+4x+3x+12=x^2+7x+12, checking every coefficient.

The next expression keeps the middle coefficient 7 but changes the constant to a negative value. The two bracket constants must now have opposite signs, while their sum is still positive.

Worked example 2: opposite signs

Question. Factorise x2+7x−18x^2+7x-18.

Plan. Find a signed pair with product −18-18 and sum 7.

9×(−2)=−189+(−2)=7x2+7x−18=(x+9)(x−2)\begin{aligned} 9\times(-2)&=-18\\ 9+(-2)&=7\\ x^2+7x-18&=(x+9)(x-2) \end{aligned}

Answer. (x+9)(x−2)(x+9)(x-2).

Check. (x+9)(x−2)=x2−2x+9x−18=x2+7x−18(x+9)(x-2)=x^2-2x+9x-18=x^2+7x-18.

For x2−7x+12x^2-7x+12, both constants would be negative: −3−4=−7-3-4=-7 and (−3)(−4)=12(-3)(-4)=12. Check the sum as well as the product. Some quadratics, such as x2+x+1x^2+x+1, have no factorisation into brackets with integer constants; do not force an unsuitable pair.

Watch out

(x+2)(x+6)(x+2)(x+6) has constant 12, but its middle term is 8x8x, not 7x7x. Matching only the constant does not establish a factorisation.

Use the two clues together.

Non-monic quadratics

When the coefficient of x2x^2 is not 1, a simple pair of constants is no longer enough. Splitting the middle term lets you recover the four terms that came from expanding two brackets.

For ax2+bx+cax^2+bx+c, look for numbers mm and nn with mn=acmn=ac and m+n=bm+n=b. Replace bxbx by mx+nxmx+nx, then factorise the two pairs. The condition mn=acmn=ac makes the same bracket appear in both pairs. You may instead find the brackets directly, provided their expansion matches all three terms.

Worked example 3: splitting a positive middle term

Question. Factorise 6x2+11x+36x^2+11x+3.

Plan. Split 11 into two numbers with product 6×3=186\times3=18, then extract common brackets.

6x2+11x+3=6x2+9x+2x+39 and 2 add to 11=3x(2x+3)+1(2x+3)factorise each pair=(3x+1)(2x+3)extract the whole shared bracket\begin{aligned} 6x^2+11x+3&=6x^2+9x+2x+3 &&\text{9 and 2 add to 11}\\ &=3x(2x+3)+1(2x+3) &&\text{factorise each pair}\\ &=(3x+1)(2x+3) &&\text{extract the whole shared bracket} \end{aligned}

Answer. (3x+1)(2x+3)(3x+1)(2x+3).

Check. Expanding gives 6x2+9x+2x+3=6x2+11x+36x^2+9x+2x+3=6x^2+11x+3.

Change only the sign of the middle term. The product of the splitting numbers stays positive, so both splitting numbers must now be negative.

Worked example 4: splitting a negative middle term

Question. Factorise 6x2−11x+36x^2-11x+3.

Plan. Use −9-9 and −2-2, whose product is 18 and sum is −11-11.

6x2−11x+3=6x2−9x−2x+3=3x(2x−3)−1(2x−3)extract a negative from the second pair=(3x−1)(2x−3)\begin{aligned} 6x^2-11x+3&=6x^2-9x-2x+3\\ &=3x(2x-3)-1(2x-3) &&\text{extract a negative from the second pair}\\ &=(3x-1)(2x-3) \end{aligned}

Answer. (3x−1)(2x−3)(3x-1)(2x-3).

Check. (3x−1)(2x−3)=6x2−9x−2x+3=6x2−11x+3(3x-1)(2x-3)=6x^2-9x-2x+3=6x^2-11x+3.

The final line uses the same distributive law as uv+wv=(u+w)vuv+wv=(u+w)v. Treat (2x−3)(2x-3) as one shared factor, even though it contains two terms.

Watch out

The splitting product is acac, not just cc. Also, −2x+3=−(2x−3)-2x+3=-(2x-3): taking out −1-1 changes both signs inside the bracket.

Write the split before grouping if you need to keep the signs visible.

Common factors first

A cubic expression can hide a quadratic. If every term contains xx, extract it before looking for two brackets. Include any numerical common factor at the same time: “factorise” means factorise fully.

For example, x3+bx2+cx=x(x2+bx+c)x^3+bx^2+cx=x(x^2+bx+c). The outside xx remains part of the answer. Extracting it is an identity, not division of an equation: no value, including x=0x=0, is lost.

Worked example 5: a cubic with a shared letter

Question. Factorise x3+5x2+6xx^3+5x^2+6x.

Plan. Extract xx, then factorise the remaining monic quadratic.

x3+5x2+6x=x(x2+5x+6)every term contains x=x(x+2)(x+3)2 and 3 multiply to 6 and add to 5\begin{aligned} x^3+5x^2+6x&=x(x^2+5x+6) &&\text{every term contains }x\\ &=x(x+2)(x+3) &&\text{2 and 3 multiply to 6 and add to 5} \end{aligned}

Answer. x(x+2)(x+3)x(x+2)(x+3).

Check. (x+2)(x+3)=x2+5x+6(x+2)(x+3)=x^2+5x+6; multiplying by xx returns x3+5x2+6xx^3+5x^2+6x. At x=0x=0, both forms are also zero.

The next expression changes only by an overall factor of 2. Remove that numerical factor together with the letter so that the same quadratic appears.

Worked example 6: a numerical factor as well

Question. Factorise 2x3+10x2+12x2x^3+10x^2+12x.

Plan. Extract the full common factor 2x2x, then reuse the quadratic pattern.

2x3+10x2+12x=2x(x2+5x+6)=2x(x+2)(x+3)\begin{aligned} 2x^3+10x^2+12x&=2x(x^2+5x+6)\\ &=2x(x+2)(x+3) \end{aligned}

Answer. 2x(x+2)(x+3)2x(x+2)(x+3).

Check. Re-expansion gives 2x(x2+5x+6)=2x3+10x2+12x2x(x^2+5x+6)=2x^3+10x^2+12x. The leading term is 2x×x×x=2x32x\times x\times x=2x^3, so the outside factor has been kept.

If the remaining quadratic is non-monic, apply the splitting method from the preceding part. The order is still common factor first, quadratic second, expansion check last.

Watch out

2x(x2+5x+6)2x(x^2+5x+6) is correct but unfinished. (x+2)(x+3)(x+2)(x+3) loses the factor 2x2x. The full answer needs all three factors and the numerical multiplier.

Finish every factorisation you can see.

Quick Recap

  • For a monic quadratic, bracket constants have the required sum and product.
  • For a non-monic quadratic, split the middle coefficient using product acac, then group.
  • Remove a full common factor before factorising a remaining quadratic.
  • Expand the proposed factors to check every original term.

The opening expression is (x+3)(x+4)(x+3)(x+4): 3 and 4 multiply to 12 and add to 7.