P1 1.10 - Expanding and factorising

P1 1.10 - Expanding and factorising

Move between products and sums without changing an expression's value. Expanding reveals coefficients; factorising reveals hidden factors, which will later make equations easier to solve.

Expanding products and collecting terms

How can x3+2x2−9x−18x^3+2x^2-9x-18 be written as three simple brackets? We will build up to this by treating expansion and factorisation as opposite directions along the same route.

Multiplication distributes across addition: a(b+c)=ab+aca(b+c)=ab+ac. For two brackets, every term in the first multiplies every term in the second. Then collect terms with identical powers of xx.

Polynomial

A polynomial in xx is a sum of constant multiples of nonnegative integer powers of xx. Its degree is the highest power with a nonzero coefficient. For example, 4x3−x+24x^3-x+2 has degree 33; 1/x1/x is not a polynomial in xx.

The terms 3x23x^2 and −5x2-5x^2 are like terms, giving −2x2-2x^2. The terms 3x23x^2 and 5x5x have different powers, so they cannot be collected. When f(x)=x2+5x+6f(x)=x^2+5x+6, the notation f(2)f(2) means substitute 22 for every xx.

Worked example 1: two sums

Expand and simplify (x+3)(x+2)(x+3)(x+2).

Plan. Distribute xx and then 33 across the second bracket.

(x+3)(x+2)=x2+2x+3x+6form all four products=x2+5x+6collect the two linear terms\begin{aligned} (x+3)(x+2)&=x^2+2x+3x+6 &&\text{form all four products}\\ &=x^2+5x+6 &&\text{collect the two linear terms} \end{aligned}

Answer. x2+5x+6x^2+5x+6.

Check. At x=2x=2, the product is 5×4=205\times4=20; the expanded form gives 4+10+6=204+10+6=20.

Change only the sign of the 22 in the second bracket. Both products involving that term change sign.

Worked example 2: a negative term

Expand and simplify (x+3)(x−2)(x+3)(x-2).

Plan. Keep the minus sign attached to −2-2 in each multiplication.

(x+3)(x−2)=x2−2x+3x−6distribute every term=x2+x−6collect like powers\begin{aligned} (x+3)(x-2)&=x^2-2x+3x-6 &&\text{distribute every term}\\ &=x^2+x-6 &&\text{collect like powers} \end{aligned}

Answer. x2+x−6x^2+x-6.

Check. At x=1x=1, the original gives 4×(−1)=−44\times(-1)=-4, and 1+1−6=−41+1-6=-4.

The same distribution works with coefficients: (2x)(3x)=6x2(2x)(3x)=6x^2. With a bracket outside a subtraction, its minus sign affects every term inside.

Watch out

4x−(x2−3x)=4x−x2+3x=−x2+7x4x-(x^2-3x)=4x-x^2+3x=-x^2+7x. Removing a bracket after a minus sign changes every sign, not only the first.

Use all four products before collecting terms.

Common factors in cubics

Factorising reverses expansion: it writes a sum as a product, starting with anything every term shares.

Factorising

To factorise an expression is to write it as a product of factors. A common factor divides every term. The highest common numerical factor and the smallest power present in every term give the highest common monomial factor.

For 12x3+8x212x^3+8x^2, both terms contain 4x24x^2, giving 4x2(3x+2)4x^2(3x+2). Dividing each term by the proposed outside factor tells you exactly what belongs inside. A missing constant term does not mean the expression is quadratic: its highest power still determines its degree.

Worked example 3: two terms in a cubic

Take out the highest common factor of 6x3+9x26x^3+9x^2.

Plan. The numerical common factor is 33, and both terms contain x2x^2.

6x3+9x2=3x2(6x33x2+9x23x2)=3x2(2x+3)divide each term by 3x2\begin{aligned} 6x^3+9x^2&=3x^2\left(\frac{6x^3}{3x^2}+\frac{9x^2}{3x^2}\right)\\ &=3x^2(2x+3) &&\text{divide each term by }3x^2 \end{aligned}

The divisions here identify the factorisation; the final polynomial identity also holds at x=0x=0.

Answer. 3x2(2x+3)3x^2(2x+3).

Check. Expanding gives 3x2×2x+3x2×3=6x3+9x23x^2\times2x+3x^2\times3=6x^3+9x^2.

Change the expression by adding one linear term. That lowers the power of xx shared by every term.

Worked example 4: the smallest power controls the factor

Take out the highest common factor of 6x3+9x2−3x6x^3+9x^2-3x.

Plan. Use 3x3x, because the final term contains only one factor of xx.

6x3+9x2−3x=3x(2x2+3x−1)divide each term by 3x\begin{aligned} 6x^3+9x^2-3x &=3x(2x^2+3x-1) &&\text{divide each term by }3x \end{aligned}

Answer. 3x(2x2+3x−1)3x(2x^2+3x-1).

Check. At x=−1x=-1, the original is −6+9+3=6-6+9+3=6, while the product is −3(2−3−1)=6-3(2-3-1)=6.

The final −1-1 is needed: multiplying it by 3x3x restores the original −3x-3x.

Watch out

x3+4x2+3xx^3+4x^2+3x has common factor xx, giving x(x2+4x+3)x(x^2+4x+3). Taking x2x^2 outside would leave a term 3/x3/x, so it has not found a polynomial common factor of all three terms.

For now, stop once the highest common factor is outside; the next part teaches how to factorise the quadratic that may remain.

Quadratic factorisation

For a quadratic with leading coefficient 11, the two bracket numbers must add to the coefficient of xx and multiply to the constant.

Expanding (x+p)(x+q)(x+p)(x+q) explains why:

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

If the constant is negative, the numbers have opposite signs. If the constant is positive, they have the same sign, chosen to give the required sum. Not every quadratic factorises with integer bracket numbers.

Worked example 5: leading coefficient one

Factorise x2+x−6x^2+x-6.

Plan. Find two integers whose product is −6-6 and whose sum is 11.

3×(−2)=−6,3+(−2)=1,3\times(-2)=-6,\qquad 3+(-2)=1,

so

x2+x−6=(x+3)(x−2).x^2+x-6=(x+3)(x-2).

Answer. (x+3)(x−2)(x+3)(x-2).

Check. The cross-products give −2x+3x=x-2x+3x=x, and the constant product is −6-6.

Change only the coefficient of x2x^2 from 11 to 22. The bracket search must now account for that coefficient.

Worked example 6: leading coefficient two

Factorise 2x2+x−62x^2+x-6.

Plan. Split the middle term using two numbers with product 2×(−6)=−122\times(-6)=-12 and sum 11, then factorise by grouping.

2x2+x−6=2x2+4x−3x−6split x as 4x−3x=2x(x+2)−3(x+2)factor each pair=(2x−3)(x+2)take out the shared bracket\begin{aligned} 2x^2+x-6&=2x^2+4x-3x-6 &&\text{split }x\text{ as }4x-3x\\ &=2x(x+2)-3(x+2) &&\text{factor each pair}\\ &=(2x-3)(x+2) &&\text{take out the shared bracket} \end{aligned}

Why split this way? It creates the same factor x+2x+2 in both pairs, so distribution can be reversed again.

Answer. (2x−3)(x+2)(2x-3)(x+2).

Check. Re-expanding gives 2x2+4x−3x−6=2x2+x−62x^2+4x-3x-6=2x^2+x-6.

Use the product of the leading coefficient and the constant when splitting a non-monic quadratic.

Watch out

For 2x2+x−62x^2+x-6, choosing numbers that merely multiply to −6-6 ignores the leading 22. Every proposed factorisation should be expanded to check all three coefficients.

The middle term can be split in either order.

Choosing a factorisation method

Look at the structure before searching for bracket numbers: a common factor, two squares or a repeated bracket can make a long expression much simpler.

Difference of two squares

A2−B2=(A−B)(A+B).A^2-B^2=(A-B)(A+B).

The two cross-products cancel. For example, 9x2−16=(3x−4)(3x+4)9x^2-16=(3x-4)(3x+4).

Not in the formula booklet: learn it.

First remove any common factor. A two-term difference of squares then has an immediate factorisation; a three-term quadratic suggests the method just taught. With four terms, try grouping pairs to create a shared bracket. Continue until no further factorisation of the kind sought is possible.

Worked example 7: a cubic with no common power

Factorise fully x3+2x2−9x−18x^3+2x^2-9x-18.

Plan. Group the first two and last two terms to expose x+2x+2, then use a difference of squares.

x3+2x2−9x−18=x2(x+2)−9(x+2)factor each pair=(x+2)(x2−9)take out the shared bracket=(x+2)(x−3)(x+3)factor x2−32\begin{aligned} x^3+2x^2-9x-18 &=x^2(x+2)-9(x+2) &&\text{factor each pair}\\ &=(x+2)(x^2-9) &&\text{take out the shared bracket}\\ &=(x+2)(x-3)(x+3) &&\text{factor }x^2-3^2 \end{aligned}

Answer. (x+2)(x−3)(x+3)(x+2)(x-3)(x+3), answering the opening question.

Check. At x=1x=1, the cubic gives 1+2−9−18=−241+2-9-18=-24; the product gives 3×(−2)×4=−243\times(-2)\times4=-24. Expanding (x+2)(x2−9)(x+2)(x^2-9) restores all four original terms.

Change the square number in the grouped expression from 99 to 44. The same strategy works, but two factors now coincide.

Worked example 8: a repeated factor

Factorise fully x3+2x2−4x−8x^3+2x^2-4x-8.

Plan. Expose x+2x+2 by grouping, then factorise x2−4x^2-4.

x3+2x2−4x−8=x2(x+2)−4(x+2)=(x+2)(x2−4)=(x+2)(x−2)(x+2)=(x+2)2(x−2).\begin{aligned} x^3+2x^2-4x-8 &=x^2(x+2)-4(x+2)\\ &=(x+2)(x^2-4)\\ &=(x+2)(x-2)(x+2)\\ &=(x+2)^2(x-2). \end{aligned}

Answer. (x+2)2(x−2)(x+2)^2(x-2).

Check. At x=0x=0, both forms give −8-8. Expanding the last form gives (x2+4x+4)(x−2)=x3+2x2−4x−8(x^2+4x+4)(x-2)=x^3+2x^2-4x-8.

The repeated factor has not disappeared; its power records both copies.

Watch out

A2+B2A^2+B^2 does not use the difference-of-squares identity. For example, (x−3)(x+3)=x2−9(x-3)(x+3)=x^2-9, not x2+9x^2+9.

Use grouping, then check whether the remaining quadratic factor can be split further.

Quick Recap

  • Expand by multiplying every term, then collect identical powers.
  • Factorise by looking first for a factor shared by every term.
  • A quadratic's middle coefficient comes from its two cross-products.
  • Grouping can reduce a cubic to a linear factor times a quadratic; keep factorising when possible.

The opening cubic is (x+2)(x−3)(x+3)(x+2)(x-3)(x+3): grouping revealed the brackets.