Factorising

The highest common factor must include the largest number and the highest power of x that divide both terms.

1. The highest common factor of 28 and 12 is 4
2. The highest power of x in both terms is x^2
3. So take out 4x^2: 28x^3\div4x^2=7x and 12x^2\div4x^2=3
4. 4x(7x^2-3x) and 2x^2(14x-6) are correct but not fully factorised: a common factor is left inside the bracket
5. 4x^2(7x-12) expands to 28x^3-48x^2
6. Answer: 4x^2(7x-3)

Both terms are square numbers or squares of a term, and they are subtracted.

1. 121=11^2 and 9k^2=(3k)^2
2. This is a difference of two squares: a^2-b^2=(a-b)(a+b)
3. Answer: (11-3k)(11+3k)

Find the highest common factor of 6, 9 and 15, then the lowest power of p and of q that appears in every term.

1. The highest common factor of 6, 9 and 15 is 3
2. Every term contains at least p^2 and at least q^2
3. So the highest common factor is 3p^2q^2
4. 6p^3q^2\div3p^2q^2=2p, \;9p^2q^3\div3p^2q^2=3q, \;15p^2q^2\div3p^2q^2=5
5. Answer: 3p^2q^2(2p+3q-5)

Take care with -2x\times(-4) when you expand the second bracket.

1. x(5x+2)=5x^2+2x
2. -2x(x-4)=-2x^2+8x
3. Collect like terms: 5x^2-2x^2+2x+8x=3x^2+10x
4. The common factor is x
5. Answer: x(3x+10)

Look for two numbers that multiply to 3\times(-10)=-30 and add to -13.

1. 3\times(-10)=-30: the two numbers that multiply to -30 and add to -13 are -15 and 2
2. Split the middle term: 3x^2-15x+2x-10
3. =3x(x-5)+2(x-5)
4. Answer: (3x+2)(x-5)

Take out the highest common factor of the numbers and the letters first, then look at what is left inside the bracket.

1. The highest common factor of 45a^2b and 20b^3 is 5b
2. 45a^2b-20b^3=5b(9a^2-4b^2)
3. 9a^2-4b^2=(3a)^2-(2b)^2 is a difference of two squares
4. 9a^2-4b^2=(3a-2b)(3a+2b)
5. Answer: 5b(3a-2b)(3a+2b)

a^2-b^2 can be written as (a-b)(a+b), and here a+b is a very easy number.

1. 5.37^2-4.63^2=(5.37-4.63)(5.37+4.63)
2. =0.74\times10
3. Answer: 7.4

Take a common factor out of the first two terms and out of the last two terms, so that the same bracket appears twice.

1. First pair: 2ab-6a=2a(b-3)
2. Second pair: -b+3=-1(b-3)
3. So 2ab-6a-b+3=2a(b-3)-1(b-3)
4. (b-3) is a common factor
5. Answer: (2a-1)(b-3)

Treat it like a quadratic in a: look for two numbers that multiply to 4\times(-15)=-60 and add to -4.

1. 4\times(-15)=-60: the two numbers that multiply to -60 and add to -4 are -10 and 6
2. Split the middle term: 4a^2-10ab+6ab-15b^2
3. =2a(2a-5b)+3b(2a-5b)
4. Answer: (2a-5b)(2a+3b)

Expand the right-hand side and compare the constant terms first to find c.

1. Constant terms: -3\times c=-15, so c=5
2. Expand (4x-3)(2x+5)=8x^2+20x-6x-15
3. =8x^2+14x-15
4. Compare the x terms: k=14

Take out the highest common factor first, then see whether the quadratic left in the bracket factorises.

1. The highest common factor is 5x
2. 10x^3-35x^2+15x=5x(2x^2-7x+3)
3. 2\times3=6: the numbers that multiply to 6 and add to -7 are -6 and -1
4. 2x^2-6x-x+3=2x(x-3)-1(x-3)=(2x-1)(x-3)
5. Answer: 5x(2x-1)(x-3)

Rearrange so that one side is 0 before you factorise.

1. Rearrange: 6x^2-7x-3=0
2. 6\times(-3)=-18: the numbers that multiply to -18 and add to -7 are -9 and 2
3. 6x^2-9x+2x-3=3x(2x-3)+1(2x-3)
4. (3x+1)(2x-3)=0
5. 3x+1=0 gives x=-\frac13 and 2x-3=0 gives x=\frac32
6. Answer: x=-\frac13 or x=\frac32

Don't divide both sides by x: rearrange to 25x^3-4x=0 and take out the common factor instead.

1. Rearrange: 25x^3-4x=0
2. Take out the common factor x: x(25x^2-4)=0
3. Difference of two squares: x(5x-2)(5x+2)=0
4. So x=0, 5x-2=0 or 5x+2=0
5. Answer: x=0, x=\frac25 or x=-\frac25
6. (Dividing by x at the start loses the solution x=0.)

This is A^2-B^2 with A=x+6 and B=3x-2, so use (A-B)(A+B) and keep the brackets when you subtract.

1. Difference of two squares: \left[(x+6)-(3x-2)\right]\left[(x+6)+(3x-2)\right]
2. First bracket: x+6-3x+2=8-2x
3. Second bracket: x+6+3x-2=4x+4
4. Take out common factors: 8-2x=2(4-x) and 4x+4=4(x+1)
5. 2\times4=8
6. Answer: 8(4-x)(x+1)

(x+4) is a factor of both terms, so take it out first and then look at what is left.

1. Take out the common factor (x+4): (x+4)\left[(x+4)^2-9\right]
2. (x+4)^2-9 is a difference of two squares: \left[(x+4)-3\right]\left[(x+4)+3\right]
3. =(x+1)(x+7)
4. Answer: (x+4)(x+1)(x+7)

Write 4 and 8 as powers of 2, so both sides are a single power of 2, then equate the powers.

1. 4^{x^2}=(2^2)^{x^2}=2^{2x^2}
2. Left-hand side: \dfrac{2^{2x^2}}{2^x}=2^{2x^2-x}
3. Right-hand side: 8^{x+2}=(2^3)^{x+2}=2^{3x+6}
4. Equate the powers: 2x^2-x=3x+6
5. 2x^2-4x-6=0, so x^2-2x-3=0
6. (x-3)(x+1)=0
7. Answer: x=3 or x=-1

In (b), let y=\sin x and rearrange so the right-hand side is 0. Answer part (a) before attempting part (b).

Part (a)

1. 2\times(-1)=-2: the numbers that multiply to -2 and add to 1 are 2 and -1
2. 2y^2+2y-y-1=2y(y+1)-1(y+1)
3. Answer: (2y-1)(y+1)

Part (b)

1. Rearrange: 2\sin^2x+\sin x-1=0
2. This is part (a) with y=\sin x: (2\sin x-1)(\sin x+1)=0
3. \sin x=\frac12 gives x=30^\circ or x=180^\circ-30^\circ=150^\circ
4. \sin x=-1 gives x=270^\circ
5. Answer: x=30^\circ, 150^\circ or 270^\circ

Differentiate, set the derivative equal to 0 and take out the highest common factor; don't divide by x. Answer part (a) before attempting part (b).

Part (a)

1. \dfrac{dy}{dx}=12x^3+12x^2-24x
2. At a stationary point \dfrac{dy}{dx}=0: 12x^3+12x^2-24x=0
3. Take out the common factor 12x: 12x(x^2+x-2)=0
4. 12x(x+2)(x-1)=0
5. Answer: x=-2, x=0 or x=1

Part (b)

1. \dfrac{d^2y}{dx^2}=36x^2+24x-24
2. At x=-2: 144-48-24=72>0, a minimum
3. At x=0: -24<0, a maximum
4. At x=1: 36+24-24=36>0, a minimum
5. When x=0, y=5
6. Answer: (0,\ 5)

The uncovered area is the area of the card minus the area of the picture, and that is a difference of two squares. Answer part (a) before attempting part (b).

Part (a)

1. A=(2x+5)^2-(x+3)^2
2. Difference of two squares: \left[(2x+5)-(x+3)\right]\left[(2x+5)+(x+3)\right]
3. =(x+2)(3x+8)
4. Answer: A=(x+2)(3x+8)

Part (b)

1. (x+2)(3x+8)<120
2. Expand: 3x^2+14x+16<120
3. 3x^2+14x-104<0
4. Factorise: (x-4)(3x+26)<0
5. The critical values are x=4 and x=-\frac{26}{3}, so -\frac{26}{3}<x<4
6. x is positive, so the lower limit is 0
7. Answer: 0<x<4

In both parts, look for a factor common to every term before doing anything else; in (b) don't cancel it.

Part (a)

1. Numerator: take out 3x: 12x^3-27x=3x(4x^2-9)
2. 4x^2-9 is a difference of two squares: 3x(2x-3)(2x+3)
3. Denominator: 6x^2-5x-6=(2x-3)(3x+2)
4. Cancel the common factor (2x-3)
5. Answer: \dfrac{3x(2x+3)}{3x+2}

Part (b)

1. Do not divide both sides by (2x-3)^2: that loses a solution
2. Rearrange: x(2x-3)^3-5(2x-3)^2=0
3. Take out the common factor (2x-3)^2: (2x-3)^2\left[x(2x-3)-5\right]=0
4. Simplify the bracket: (2x-3)^2(2x^2-3x-5)=0
5. Factorise: (2x-3)^2(2x-5)(x+1)=0
6. Answer: x=\frac32, x=\frac52, x=-1