Factorising Harder Quadratics

Look for factors of the constant that give the required cross products.

1. The 2x^2 can only come from 2x\times x, so one bracket starts with 2x and the other with x.
2. The +3 and the +7x are both positive, so the numbers in the brackets are 1 and 3, both positive.
3. Try (2x+1)(x+3): the cross products are 2x\times3=6x and 1\times x=x, which add to 7x.
4. So 2x^2+7x+3=(2x+1)(x+3). (The other way round, (2x+3)(x+1), gives 5x in the middle.)

The first terms of the brackets multiply to the squared term.

1. The 3x^2 comes from 3x\times x, so one bracket starts with 3x and the other with x; the numbers multiply to +5: they are 5 and 1.
2. Try (3x+5)(x+1): the cross products are 3x\times1=3x and 5\times x=5x, which add to 8x.
3. So 3x^2+8x+5=(3x+5)(x+1).

A positive constant and a negative middle term require two negative constants in the brackets.

1. Multiply the x^2 coefficient by the number term: 2\times4=8. Find two numbers that multiply to 8 and add to -9: they are -8 and -1.
2. Split the middle term using them: 2x^2-8x-x+4.
3. Factorise each pair: 2x(x-4)-1(x-4).
4. Take out the common bracket (x-4): (2x-1)(x-4).

The constants in the brackets have opposite signs.

1. Multiply the x^2 coefficient by the number term: 3\times(-2)=-6. Two numbers that multiply to -6 and add to +1 are 3 and -2.
2. Split the middle term: 3x^2+3x-2x-2.
3. Factorise each pair: 3x(x+1)-2(x+1).
4. Take out the common bracket (x+1): (3x-2)(x+1).

Recognise a difference of two squares.

1. Both terms are squares and one is subtracted from the other, so use the difference of two squares: a^2-b^2=(a-b)(a+b).
2. Here 4x^2=(2x)^2 and 25=5^2, so a=2x and b=5.
3. So 4x^2-25=(2x-5)(2x+5).

Choose two numbers with product equal to the leading coefficient times the constant.

1. Multiply the x^2 coefficient by the number term: 6\times6=36. Two numbers that multiply to 36 and add to 13 are 9 and 4.
2. Split the middle term using them: 6x^2+9x+4x+6.
3. Factorise each pair: 3x(2x+3)+2(2x+3).
4. Take out the common bracket (2x+3): (3x+2)(2x+3).

Split the middle term into two terms with coefficients whose product is negative.

1. Multiply the x^2 coefficient by the number term: 6\times(-6)=-36. Two numbers that multiply to -36 and add to -5 are -9 and 4.
2. Split the middle term using them: 6x^2-9x+4x-6.
3. Factorise each pair: 3x(2x-3)+2(2x-3).
4. Take out the common bracket (2x-3): (3x+2)(2x-3).

The product needed for splitting the middle term is positive, but the sum is negative.

1. Multiply the x^2 coefficient by the number term: 8\times3=24. Two numbers that multiply to 24 and add to -10 are -6 and -4.
2. Split the middle term using them: 8x^2-4x-6x+3.
3. Factorise each pair: 4x(2x-1)-3(2x-1). Taking out -3 turns -6x+3 into -3(2x-1).
4. Take out the common bracket (2x-1): (4x-3)(2x-1).

Check whether the expression is the square of a bracket.

1. The first and last terms are squares: 9x^2=(3x)^2 and 4=2^2.
2. The middle term is 2\times3x\times2=12x, which matches, so this is a perfect square: (3x+2)(3x+2).
3. So 9x^2+12x+4=(3x+2)^2.

Take out a common numerical factor before factorising the quadratic.

1. The terms 6x^2, 21x and -12 have a common factor of 3: 6x^2+21x-12=3(2x^2+7x-4).
2. For 2x^2+7x-4: 2\times(-4)=-8, and two numbers that multiply to -8 and add to 7 are 8 and -1.
3. Split and factorise in pairs: 2x^2+8x-x-4=2x(x+4)-1(x+4)=(2x-1)(x+4).
4. Remember the 3 taken out at the start: 6x^2+21x-12=3(2x-1)(x+4).

Keep the second letter with the constants when choosing the brackets.

1. Treat q like a number. Multiply the outer coefficients: 12\times(-2)=-24. Two numbers that multiply to -24 and add to 5 are 8 and -3.
2. Split the middle term: 12p^2+8pq-3pq-2q^2.
3. Factorise each pair: 4p(3p+2q)-q(3p+2q).
4. Take out the common bracket: (3p+2q)(4p-q).

The pair must add to the middle coefficient and multiply to the leading coefficient times the constant.

1. The required product is 5\times(-6)=-30.
2. The pair -15 and 2 has product -30 and sum -13.
3. So the useful split is -15t+2t.

Factorise the quadratic, then set each bracket equal to zero.

1. Factorise the left-hand side. 6\times(-2)=-12, and two numbers that multiply to -12 and add to 1 are 4 and -3.
2. 6x^2+4x-3x-2=2x(3x+2)-1(3x+2)=(3x+2)(2x-1), so (3x+2)(2x-1)=0.
3. If two brackets multiply to 0, one of them must be 0: 3x+2=0 or 2x-1=0.
4. So x=-\frac{2}{3} or x=\frac{1}{2}.

Factorise the area expression to identify the other length factor.

1. Area of a rectangle = length \times width, so width = area \div length.
2. Factorise the area: 6\times(-3)=-18, and 9 and -2 multiply to -18 and add to 7. So 6x^2+9x-2x-3=3x(2x+3)-1(2x+3)=(3x-1)(2x+3).
3. One factor is the length (3x-1), so the other factor is the width.
4. The width is (2x+3) cm.

Factorise the numerator before cancelling a common factor.

1. Factorise the numerator: 4\times(-3)=-12, and 6 and -2 multiply to -12 and add to 4.
2. 4x^2+6x-2x-3=2x(2x+3)-1(2x+3)=(2x-1)(2x+3).
3. So the fraction is \dfrac{(2x-1)(2x+3)}{2x+3}. Cancel the common factor (2x+3), which is not zero because x\ne-\frac{3}{2}.
4. This leaves 2x-1.

A negative leading coefficient means the leading terms in the brackets have opposite signs.

1. Multiply the outer coefficients: (-6)\times3=-18. Two numbers that multiply to -18 and add to 17 are 18 and -1.
2. Split the middle term: -6x^2+18x-x+3.
3. Factorise each pair: 6x(-x+3)+1(-x+3), that is 6x(3-x)+1(3-x).
4. Take out the common bracket: (3-x)(6x+1). (This can also be written -(x-3)(6x+1).)

This is a difference of two squares: A^2-B^2=(A-B)(A+B).

1. Use the difference of two squares, A^2-B^2=(A-B)(A+B), with A=3x+5 and B=x+1.
2. A-B=(3x+5)-(x+1)=2x+4 and A+B=(3x+5)+(x+1)=4x+6.
3. So the expression is (2x+4)(4x+6).
4. Each bracket has a common factor of 2: 2x+4=2(x+2) and 4x+6=2(2x+3).
5. Fully factorised: 2\times2\times(x+2)(2x+3)=4(x+2)(2x+3).

First put the equation in a form with zero on one side.

1. To solve by factorising, first get 0 on one side: subtract 5 to get 12x^2-17x-5=0.
2. 12\times(-5)=-60. Two numbers that multiply to -60 and add to -17 are -20 and 3.
3. 12x^2-20x+3x-5=4x(3x-5)+1(3x-5)=(3x-5)(4x+1), so (3x-5)(4x+1)=0.
4. Either 3x-5=0, giving x=\frac{5}{3}, or 4x+1=0, giving x=-\frac{1}{4}.

Split the mixed term into two terms whose coefficients have product equal to the leading and final coefficients multiplied.

1. Treat b like a number. Multiply the outer coefficients: 15\times(-8)=-120. Two numbers that multiply to -120 and add to -2 are 10 and -12.
2. Split the middle term: 15a^2+10ab-12ab-8b^2.
3. Factorise each pair: 5a(3a+2b)-4b(3a+2b).
4. Take out the common bracket: (5a-4b)(3a+2b).

Factorise to find both solutions, then subtract the negative one carefully.

1. Factorise: 4\times(-15)=-60, and 10 and -6 multiply to -60 and add to 4. So 4x^2+10x-6x-15=2x(2x+5)-3(2x+5)=(2x-3)(2x+5).
2. From (2x-3)(2x+5)=0: 2x-3=0 gives x=\frac{3}{2}, and 2x+5=0 gives x=-\frac{5}{2}.
3. The positive solution is v=\frac{3}{2} and the negative solution is u=-\frac{5}{2}.
4. v-u=\frac{3}{2}-\left(-\frac{5}{2}\right)=\frac{3}{2}+\frac{5}{2}=4.