Factor Theorem

The factor theorem says \mathrm{f}(a)=0 means x=a is a solution, so think about which bracket equals zero when x=-3.

1. \mathrm{f}(-3)=0 means x=-3 is a solution of \mathrm{f}(x)=0
2. The bracket that is zero when x=-3 is (x+3)
3. So (x+3) is a factor
4. (x-3) is not a factor: \mathrm{f}(3)=27+9-15+3=24, which is not 0
5. Answer: (x+3) is a factor of \mathrm{f}(x)

Put -2 in brackets every time you substitute it: (-2)^3 is negative but (-2)^2 is positive.

1. \mathrm{f}(-2)=2(-2)^3-5(-2)^2+4(-2)+20
2. =2(-8)-5(4)-8+20
3. =-16-20-8+20
4. So (x+2) is not a factor, because \mathrm{f}(-2)\neq0
5. Answer: -24

(ax-b) is a factor of \mathrm{f}(x) exactly when \mathrm{f}\!\left(\frac{b}{a}\right)=0, so find the value of x that makes each bracket zero and substitute it.

1. (2x-3)=0 when x=\frac32
2. \mathrm{f}\!\left(\frac32\right)=2\times\frac{27}{8}-9\times\frac94+\frac32+12=\frac{27}{4}-\frac{81}{4}+\frac{6}{4}+\frac{48}{4}=0
3. So (2x-3) is a factor.
4. Check the others are not: \mathrm{f}\!\left(-\frac32\right)=-\frac{33}{2}, \mathrm{f}(1)=6, \mathrm{f}(-4)=-264, none of which is 0.
5. Watch the sign: the factor (x+4) needs \mathrm{f}(-4)=0, not \mathrm{f}(4)=0.
6. Answer: (2x-3)

If (x-2) is a factor, the expression equals 0 when x=2.

1. (x-2) is a factor, so substituting x=2 gives 0
2. 2^5-2\times2^4+2k-6=0
3. 32-32+2k-6=0
4. 2k=6
5. Answer: k=3

(3x-1) is zero when x=\frac13, so substitute x=\frac13 and set the result equal to 0.

1. 3x-1=0 when x=\frac13
2. 3\left(\frac13\right)^3+k\left(\frac13\right)^2+5\left(\frac13\right)-2=0
3. \frac19+\frac{k}{9}+\frac53-2=0
4. Multiply by 9: 1+k+15-18=0
5. Answer: k=2

Write x^3-x^2-11x-10=(x+2)(x^2+bx+c) and match the constant terms first.

1. x^3-x^2-11x-10=(x+2)(x^2+bx+c)
2. Constant term: 2c=-10, so c=-5
3. x^2 term: b+2=-1, so b=-3
4. Check the x term: c+2b=-5-6=-11 ✓
5. Answer: x^2-3x-5

The three factors are (x-2)(x+1)(x+c): what must c be to give the constant term -10?

1. x^3+4x^2-7x-10=(x-2)(x+1)(x+c)
2. Constant terms: (-2)\times1\times c=-10
3. -2c=-10, so c=5
4. Check the x^2 term: -2+1+5=4 ✓
5. Answer: (x+5)

Write \mathrm{f}(x)=(x-4)(3x^2+bx+c) and match the x^2 term and the constant term to find b and c.

1. \mathrm{f}(x)=(x-4)(3x^2+bx+c): the 3x^2 is needed to make 3x^3
2. Constant term: -4c=4, so c=-1
3. x^2 term: b-12=-10, so b=2
4. Check the x term: c-4b=-1-8=-9 ✓
5. So \mathrm{f}(x)=(x-4)(3x^2+2x-1)
6. Factorise the quadratic: 3x^2+2x-1=(3x-1)(x+1)
7. Answer: (x-4)(3x-1)(x+1)

Write the cubic as (5x-2)(x^2+bx+c): the x^2 is needed to make 5x^3.

1. 5x^3+3x^2-12x+4=(5x-2)(x^2+bx+c)
2. Constant term: -2c=4, so c=-2
3. x^2 term: 5b-2=3, so b=1
4. Check the x term: 5c-2b=-10-2=-12 ✓
5. So the cubic is (5x-2)(x^2+x-2)
6. x^2+x-2=(x+2)(x-1)
7. Answer: (5x-2)(x+2)(x-1)

Use the factor theorem to find one factor first: try x=1,\ -1,\ 2,\ -2,\dots (factors of 12) until the cubic comes to 0.

1. Let \mathrm{f}(x)=x^3-6x^2+5x+12
2. \mathrm{f}(1)=1-6+5+12=12, so (x-1) is not a factor
3. \mathrm{f}(-1)=-1-6-5+12=0, so (x+1) is a factor
4. \mathrm{f}(x)=(x+1)(x^2+bx+12) and the x^2 term gives b+1=-6, so b=-7
5. x^2-7x+12=(x-3)(x-4)
6. (x+1)(x-3)(x-4)=0
7. Answer: x=-1, x=3, x=4

Find the quadratic factor by comparing coefficients, then look carefully at its x term.

1. \mathrm{f}(x)=(x-5)(9x^2+bx+c)
2. Constant term: -5c=20, so c=-4
3. x^2 term: b-45=-45, so b=0
4. So \mathrm{f}(x)=(x-5)(9x^2-4)
5. 9x^2-4 is a difference of two squares: (3x-2)(3x+2)
6. (x-5)(3x-2)(3x+2)=0
7. Answer: x=5, x=\frac23, x=-\frac23

Substitute x=-2 and set the result equal to 0, taking care with the sign of each term containing a.

1. (x+2) is a factor, so substituting x=-2 gives 0
2. (-2)^4+a(-2)^3+(-2)^2-2a(-2)-36=0
3. 16-8a+4+4a-36=0
4. -16-4a=0
5. Answer: a=-4

Find the quadratic factor first; it does not factorise, so use the quadratic formula and leave the answers as surds.

1. x^3-2x^2-7x+2=(x+2)(x^2+bx+1), since 2\times1=2
2. x^2 term: b+2=-2, so b=-4
3. Check the x term: 1+2b=1-8=-7 ✓
4. x+2=0 gives x=-2
5. x^2-4x+1=0: x=\dfrac{4\pm\sqrt{16-4}}{2}=\dfrac{4\pm\sqrt{12}}{2}
6. \sqrt{12}=2\sqrt3, so x=2\pm\sqrt3
7. Answer: x=-2, x=2+\sqrt3, x=2-\sqrt3

After finding the quadratic factor, factorise it: you may find the same bracket twice.

1. \mathrm{f}(x)=(x-4)(9x^2+bx+1), since -4\times1=-4
2. x^2 term: b-36=-30, so b=6
3. Check the x term: 1-4b=1-24=-23 ✓
4. 9x^2+6x+1=(3x+1)^2
5. (x-4)(3x+1)^2=0
6. (3x+1) is a repeated factor, so x=-\frac13 is a repeated solution
7. Answer: x=4, x=-\frac13 (only two different solutions)

Multiply the two given factors together first, then compare coefficients with (x^2-1)(x^2+bx+c).

1. (x-1)(x+1)=x^2-1
2. So \mathrm{f}(x)=(x^2-1)(x^2+bx+c)
3. x^3 term: b=2
4. Constant term: -c=4, so c=-4
5. Check the x term: -b=-2 ✓
6. Answer: \mathrm{g}(x)=x^2+2x-4

Write \mathrm{f}(x)=(x+2)(4x^2+bx+c) and compare coefficients, then factorise the quadratic. Answer part (a) before attempting part (b).

Part (a)

1. \mathrm{f}(x)=(x+2)(4x^2+bx+9), since 2\times9=18
2. x^2 term: b+8=-4, so b=-12
3. Check the x term: 9+2b=9-24=-15 ✓
4. 4x^2-12x+9=(2x-3)^2
5. Answer: (x+2)(2x-3)^2

Part (b)

1. (x+2)(2x-3)^2=0 gives x=-2 or x=\frac32
2. (2x-3)^2=0 only when 2x-3=0, so it gives just x=\frac32 (not -\frac32)
3. Answer: 2, because (2x-3) is a repeated factor

Part (c)

1. The curve meets the x-axis where \mathrm{f}(x)=0: at x=-2 and x=\frac32
2. A repeated solution is where the curve touches the axis without crossing it
3. Answer: \left(\frac32,\ 0\right)

Set the two expressions for y equal to make a cubic equal to 0; the point A tells you one factor. Answer part (a) before attempting part (b).

Part (a)

1. At the points where they meet, 2x^3-x^2=13x+6
2. 2x^3-x^2-13x-6=0
3. x=-2 is a solution, so (x+2) is a factor
4. 2x^3-x^2-13x-6=(x+2)(2x^2+bx-3), and the x^2 term gives b+4=-1, so b=-5
5. 2x^2-5x-3=(2x+1)(x-3)
6. Answer: x=-\frac12 and x=3

Part (b)

1. The greatest x-coordinate is x=3
2. On the line: y=13\times3+6=45
3. Check on the curve: 2\times27-9=45 ✓
4. Answer: (3,\ 45)

For part (a), (2x+1) is zero when x=-\frac12; after that, use your value of a and compare coefficients to find the quadratic factor.

Part (a)

1. (2x+1) is a factor, so \mathrm{f}\!\left(-\frac12\right)=0
2. 2\left(-\frac18\right)+a\left(\frac14\right)-6\left(-\frac12\right)-1=0
3. -\frac14+\frac{a}{4}+3-1=0
4. Multiply by 4: -1+a+8=0
5. Answer: a=-7

Part (b)

1. \mathrm{f}(x)=2x^3-7x^2-6x-1=(2x+1)(x^2+bx+c)
2. Constant term: c=-1
3. x^2 term: 2b+1=-7, so b=-4
4. Check the x term: 2c+b=-2-4=-6 ✓
5. Answer: x^2-4x-1

Part (c)

1. (2x+1)(x^2-4x-1)=0
2. 2x+1=0 gives x=-\frac12 (don't forget this one)
3. x^2-4x-1=0 does not factorise, so use the quadratic formula:
4. x=\dfrac{4\pm\sqrt{16+4}}{2}=\dfrac{4\pm\sqrt{20}}{2}=\dfrac{4\pm2\sqrt5}{2}=2\pm\sqrt5
5. Answer: x=-\frac12, x=2+\sqrt5, x=2-\sqrt5

Use \mathrm{f}(3)=0 and \mathrm{g}(3)=0 to write two equations in p and q, then solve them simultaneously. Answer part (a) before attempting part (b).

Part (a)

1. \mathrm{f}(3)=0: 27+9p+3q+6=0, so 9p+3q=-33, which is 3p+q=-11
2. \mathrm{g}(3)=0: 27+9q+3p+24=0, so 3p+9q=-51, which is p+3q=-17
3. From the first equation, q=-11-3p
4. Substitute: p+3(-11-3p)=-17
5. p-33-9p=-17, so -8p=16
6. Answer: p=-2

Part (b)

1. q=-11-3p
2. =-11-3(-2)
3. Answer: q=-5

Part (c)

1. \mathrm{g}(x)=x^3-5x^2-2x+24
2. \mathrm{g}(x)=(x-3)(x^2+bx-8), since -3\times(-8)=24
3. x^2 term: b-3=-5, so b=-2
4. x^2-2x-8=(x-4)(x+2)
5. Answer: x=3, x=4, x=-2

Find the quadratic factor and factorise it: notice what the three brackets have in common for whole-number values of x.

Part (a)

1. \mathrm{f}(x)=(x+4)(x^2+bx+15), since 4\times15=60
2. x^2 term: b+4=12, so b=8
3. Check the x term: 15+4b=15+32=47 ✓
4. x^2+8x+15=(x+3)(x+5)
5. Answer: (x+3)(x+4)(x+5)

Part (b)

1. \mathrm{f}(n)=(n+3)(n+4)(n+5) is the product of three consecutive integers
2. One of any two consecutive integers is even, so the product is a multiple of 2
3. One of any three consecutive integers is a multiple of 3, so the product is a multiple of 3
4. The others fail: \mathrm{f}(1)=120 is even, \mathrm{f}(2)=210 is not a multiple of 4 and \mathrm{f}(3)=336 is not a multiple of 5
5. Answer: \mathrm{f}(n) is always a multiple of 6

Part (c)

1. (k+3)(k+4)(k+5)=990: three consecutive integers with product 990
2. 990=9\times10\times11
3. So k+3=9
4. Answer: k=6