Solving Quadratics by Factorising

If two numbers multiply to give 0, one of them must be 0.

Set each bracket equal to 0 and solve each one.

Ask yourself what value of x makes each bracket equal to 0.

x on its own is one of the two factors, so it can be 0.

Factorise first, then set each bracket equal to 0.

Find two numbers that multiply to give 20 and add to give 9.

1. 4\times5=20 and 4+5=9, so x^2+9x+20=(x+4)(x+5).
2. (x+4)(x+5)=0, so either x+4=0 or x+5=0.
3. So x=-4 or x=-5.

You need two numbers that multiply to give −28 and add to give −3, so one of them is negative.

1. -7\times4=-28 and -7+4=-3, so x^2-3x-28=(x-7)(x+4).
2. (x-7)(x+4)=0, so either x-7=0 or x+4=0.
3. So x=7 or x=-4.

x² − 64 is a difference of two squares, because 64 = 8².

Part (a)

1. x^2-64=x^2-8^2.
2. So x^2-64=(x+8)(x-8).

Part (b)

1. (x+8)(x-8)=0.
2. Either x+8=0 or x-8=0.
3. So x=-8 or x=8. Do not forget the negative solution.

Factorise each one, or put x = 6 and x = −6 into each equation.

1. x^2-36=(x+6)(x-6), so x^2-36=0 gives x=-6 or x=6.
2. x^2+36=0 has no solutions, because x^2+36 is always at least 36.
3. x^2-6=0 and x^2-12=0 do not work: 6^2-6=30 and 6^2-12=24, not 0.
4. So the answer is x^2-36=0.

Both terms have a factor of x. Take it out first.

1. x^2+7x=x(x+7).
2. x(x+7)=0, so either x=0 or x+7=0.
3. So x=0 or x=-7.

The middle term is +1x, so look for two numbers that multiply to give −30 and differ by 1.

1. The middle term is x, which is 1x.
2. 6\times(-5)=-30 and 6+(-5)=1, so x^2+x-30=(x+6)(x-5).
3. Either x+6=0 or x-5=0.
4. So x=-6 or x=5.

Write the terms in the usual order first: the x^2 term, then the x term, then the number.

1. In the usual order the equation is x^2+4x-21=0.
2. 7\times(-3)=-21 and 7+(-3)=4, so x^2+4x-21=(x+7)(x-3).
3. Either x+7=0 or x-3=0.
4. So x=-7 or x=3.

The two numbers in the brackets must multiply to give −24 and add to give −2.

Part (a)

1. -6\times4=-24 and -6+4=-2.
2. (x+6)(x-4) gives +2x, and (x-8)(x+3) gives -5x.
3. (x-12)(x+2) gives -10x.
4. So x^2-2x-24=(x-6)(x+4).

Part (b)

1. (x-6)(x+4)=0.
2. Either x-6=0 or x+4=0.
3. So x=6 or x=-4.

What do you get when you put x = 0 into x² − 9x?

Part (a)

1. Put x=0 into x^2-9x: 0-0=0, so x=0 is also a solution.
2. Dividing by x loses this solution.
3. Factorising keeps it: x(x-9)=0, so x=0 or x=9.

Part (b)

1. x^2-9x=x(x-9).
2. x(x-9)=0, so x=0 or x=9.

Look for two numbers that multiply to give 25 and add to give −10.

Part (a)

1. -5\times-5=25 and -5+(-5)=-10.
2. So x^2-10x+25=(x-5)(x-5), which can be written (x-5)^2.

Part (b)

1. (x-5)(x-5)=0.
2. Both brackets give x-5=0, so there is only one solution.
3. x=5.

Put Amir's values back into the equation to test them.

Part (a)

1. Test x=7: 49+14-35=28, not 0, so Amir is wrong.
2. The factorisation is (x+7)(x-5), since 7\times(-5)=-35 and 7+(-5)=2.
3. Amir has used the numbers in the brackets as the solutions without changing their signs.

Part (b)

1. (x+7)(x-5)=0.
2. Either x+7=0 or x-5=0.
3. So x=-7 or x=5.
4. Check: 25+10-35=0.

Solve the equation by factorising, then decide which solution makes sense for a number of people.

1. Find two numbers that multiply to give -56 and add to give -1: -8 and 7.
2. n^2-n-56=(n-8)(n+7)=0, so n=8 or n=-7.
3. The number of friends cannot be negative, so n=8.
4. There are 8 friends in the group.

Factorise to find two values of x, then think about which one can be a length.

Part (a)

1. Find two numbers that multiply to give -36 and add to give 5: 9 and -4.
2. x^2+5x-36=(x+9)(x-4)=0.
3. So x=-9 or x=4.

Part (b)

1. A width cannot be negative, so x=4.
2. The length is x+5=4+5=9 cm.
3. Check: 9\times4=36 cm^2.

Every point on the x-axis has a y-coordinate of 0.

Part (a)

1. Find two numbers that multiply to give 8 and add to give -6: -2 and -4.
2. x^2-6x+8=(x-2)(x-4)=0.
3. So x=2 or x=4.

Part (b)

1. On the x-axis, y=0, so the graph crosses the x-axis where x^2-6x+8=0.
2. From part (a), that is at x=2 and x=4.
3. The point with the larger x-coordinate is (4,0).

Solve each equation separately, then look for the value that appears in both lists.

1. x^2-2x-8=(x-4)(x+2)=0, so x=4 or x=-2.
2. x^2+5x+6=(x+2)(x+3)=0, so x=-2 or x=-3.
3. The value in both lists is x=-2.
4. Check: 4+4-8=0 and 4-10+6=0.