2.7 - Solving quadratics by factorisation

2.7 - Solving quadratics by factorisation

Factorisation turns a quadratic equation into a product of simpler factors. Learn to find every root, including zero and repeated roots, and check that none has been lost. Foundation uses monic quadratics; Higher also uses general factorisable quadratics.

The zero-product property

Could cancelling a letter make a correct answer disappear? The equation x2−6x=0x^2-6x=0 will show that it can. First we need the rule that keeps every possible answer safe.

You already know

Expanding multiplies out brackets; factorising reverses that process. A solution of an equation makes its two sides equal when substituted.

For example, 0×7=00\times7=0, but two nonzero real numbers cannot multiply to give zero. That small fact is the key to solving a factorised quadratic.

Zero-product property

For real numbers AA and BB, if AB=0AB=0, then A=0A=0 or B=0B=0. A root of an equation is a value that makes the original equation true. The factors do not have to be zero at the same value.

This rule needs a zero product. The equation AB=6AB=6 does not mean A=0A=0 or B=0B=0. The zero-product property is not supplied on either the Foundation or Higher formulae sheet: learn it.

Worked example 1: solving each factor

Question. Solve (x−4)(x+1)=0(x-4)(x+1)=0.

Plan. Set each factor equal to zero and solve the two linear equations.

x−4=0orx+1=0a factor must be zerox=4orx=−1solve each equation\begin{aligned} x-4&=0\quad\text{or}\quad x+1=0 &&\text{a factor must be zero}\\ x&=4\quad\text{or}\quad x=-1 &&\text{solve each equation} \end{aligned}

Why “or”? Each value separately makes the product zero; the two factors need not vanish together.

Answer. x=4x=4 or x=−1x=-1.

Check. At 44, the product is 0×5=00\times5=0. At −1-1, it is (−5)×0=0(-5)\times0=0.

Change only the sign inside the first bracket. The root from that bracket changes sign too.

Worked example 2: reading a positive bracket constant

Question. Solve (x+4)(x+1)=0(x+4)(x+1)=0.

Plan. Solve both factor equations rather than copying their constants.

x+4=0orx+1=0zero-product propertyx=−4orx=−1subtract the constants\begin{aligned} x+4&=0\quad\text{or}\quad x+1=0 &&\text{zero-product property}\\ x&=-4\quad\text{or}\quad x=-1 &&\text{subtract the constants} \end{aligned}

Why solve the brackets? The constant +4+4 becomes the root −4-4, because that is the value that makes x+4x+4 zero.

Answer. x=−4x=-4 or x=−1x=-1.

Check. The products are 0×(−3)=00\times(-3)=0 and 3×0=03\times0=0.

State all distinct roots, even when one is negative. A contextual restriction may remove a root later; an ordinary algebraic equation has no automatic positive-only restriction.

Watch out

(x+4)(x+1)=0⇒x=4,1(x+4)(x+1)=0\Rightarrow x=4,1 has the signs wrong. The correct factor equations give x=−4,−1x=-4,-1. Check by putting each proposed value into a bracket.

Now solve a new zero product.

Solving monic quadratics

A monic quadratic has coefficient 11 on x2x^2. Its factorisation can be found from a sum and a product.

Expanding (x+r)(x+s)(x+r)(x+s) gives x2+(r+s)x+rsx^2+(r+s)x+rs. The middle coefficient is the sum of the bracket constants, and the final term is their product. This explains which two conditions a factor pair must meet.

Monic factorisation

x2+bx+c=(x+r)(x+s),r+s=b,rs=cx^2+bx+c=(x+r)(x+s),\qquad r+s=b,\quad rs=c

Use the zero-product property after setting the factorisation equal to zero.
Not in either tier's formulae sheet: learn it.

If c<0c<0, the constants have opposite signs. If c>0c>0, they have the same sign, and their sum tells you which sign. Product alone is not enough.

Worked example 3: opposite-sign factors

Question. Solve x2+x−12=0x^2+x-12=0 by factorisation.

Plan. Find two numbers with product −12-12 and sum 11, then solve the factors.

4+(−3)=1,4(−3)=−12test both conditionsx2+x−12=(x+4)(x−3)write the factorsx+4=0orx−3=0zero productx=−4orx=3solve each factor\begin{aligned} 4+(-3)&=1,\quad4(-3)=-12 &&\text{test both conditions}\\ x^2+x-12&=(x+4)(x-3) &&\text{write the factors}\\ x+4&=0\quad\text{or}\quad x-3=0 &&\text{zero product}\\ x&=-4\quad\text{or}\quad x=3 &&\text{solve each factor} \end{aligned}

Why these signs? The larger magnitude must be positive to give a positive sum of 11.

Answer. x=−4x=-4 or x=3x=3.

Check. Expansion gives x2+(4−3)x−12x^2+(4-3)x-12. Substitution gives 16−4−12=016-4-12=0 and 9+3−12=09+3-12=0.

Change only the sign of the middle coefficient. The product stays negative, but the larger magnitude must now be negative.

Worked example 4: changing the middle sign

Question. Solve x2−x−12=0x^2-x-12=0 by factorisation.

Plan. Use product −12-12 and sum −1-1.

−4+3=−1,(−4)(3)=−12choose the signed pair(x−4)(x+3)=0factorisex−4=0orx+3=0zero productx=4orx=−3solve\begin{aligned} -4+3&=-1,\quad(-4)(3)=-12 &&\text{choose the signed pair}\\ (x-4)(x+3)&=0 &&\text{factorise}\\ x-4&=0\quad\text{or}\quad x+3=0 &&\text{zero product}\\ x&=4\quad\text{or}\quad x=-3 &&\text{solve} \end{aligned}

Why recheck the sum? The pair from Example 3 still has the correct product but has the wrong middle coefficient here.

Answer. x=4x=4 or x=−3x=-3.

Check. 16−4−12=016-4-12=0 and 9+3−12=09+3-12=0; expansion gives x2−x−12x^2-x-12.

Not every quadratic has convenient integer factors. Check your arithmetic and the requested method rather than forcing an unsuitable pair. Other solving methods are Higher content in the next lesson.

Watch out

x2+x−12=(x+6)(x−2)x^2+x-12=(x+6)(x-2) is wrong. The product is −12-12, but the sum is 44, not 11. Use (x+4)(x−3)(x+4)(x-3).

A factorisation question and a solving question ask for different final objects: brackets are factors; values are roots.

Continue this lesson.

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