Teach Me Maths

Factorising Quadratics

Question 12 marks

Factorise x^2+5x+6

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Find two numbers that multiply to give 6 and add to give 5.

Worked solution

Factorise x^2+5x+6

  1. 2\times3=6 and 2+3=5.
  2. So x^2+5x+6=(x+2)(x+3).
  3. Check: (x+2)(x+3)=x^2+3x+2x+6=x^2+5x+6.

Question 21 mark

Sam wants to factorise x^2+7x+12

He needs two numbers that multiply to give 12 and add to give 7

Which two numbers does he need?

Hint

Check both conditions for each pair: the product and the sum.

Worked solution

Sam wants to factorise x^2+7x+12

He needs two numbers that multiply to give 12 and add to give 7

Which two numbers does he need?

  1. 2 and 6
  2. 3 and 4
  3. 1 and 12
  4. 5 and 2
  1. 3\times4=12 and 3+4=7, so the numbers are 3 and 4.
  2. 2 and 6 multiply to 12 but add to 8; 5 and 2 add to 7 but multiply to 10.
  3. So x^2+7x+12=(x+3)(x+4).

Question 32 marks

Factorise x^2+8x+12

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

List the pairs of numbers that multiply to give 12, and choose the pair that adds to 8.

Worked solution

Factorise x^2+8x+12

  1. Pairs that multiply to 12: 1 and 12, 2 and 6, 3 and 4.
  2. Only 2+6=8.
  3. So x^2+8x+12=(x+2)(x+6).

Question 42 marks

Factorise x^2-7x+10

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

The numbers must multiply to give positive 10 but add to give negative 7, so think about their signs.

Worked solution

Factorise x^2-7x+10

  1. Both numbers must be negative: (-2)\times(-5)=10 and -2+(-5)=-7.
  2. So x^2-7x+10=(x-2)(x-5).
  3. Check: (x-2)(x-5)=x^2-5x-2x+10=x^2-7x+10.

Question 52 marks

Factorise x^2+2x-15

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

The numbers multiply to give negative 15, so one is positive and one is negative.

Worked solution

Factorise x^2+2x-15

  1. 5\times(-3)=-15 and 5+(-3)=2.
  2. So x^2+2x-15=(x+5)(x-3).
  3. Check: (x+5)(x-3)=x^2-3x+5x-15=x^2+2x-15.

Question 62 marks

Factorise x^2-4x-12

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Find two numbers that multiply to give -12 and add to give -4.

Worked solution

Factorise x^2-4x-12

  1. The numbers multiply to give -12, so one is positive and one is negative.
  2. -6\times2=-12 and -6+2=-4.
  3. So x^2-4x-12=(x-6)(x+2).
  4. Check: (x-6)(x+2)=x^2+2x-6x-12=x^2-4x-12.

Question 72 marks

Factorise x^2+8x+16

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Find two numbers that multiply to give 16 and add to give 8.

Worked solution

Factorise x^2+8x+16

  1. 4\times4=16 and 4+4=8.
  2. So x^2+8x+16=(x+4)(x+4), which can also be written (x+4)^2.

Question 82 marks

Factorise x^2-49

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

There is no x term, so the two numbers must add to give 0.

Worked solution

Factorise x^2-49

  1. The numbers multiply to give -49 and add to give 0.
  2. 7\times(-7)=-49 and 7+(-7)=0.
  3. So x^2-49=(x+7)(x-7).
  4. Check: (x+7)(x-7)=x^2-7x+7x-49=x^2-49.

Question 91 mark

Factorise x^2-9x

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Both terms have a common factor of x.

Worked solution

Factorise x^2-9x

  1. x^2=x\times x and 9x=9\times x, so x is a common factor.
  2. x^2-9x=x(x-9).

Question 101 mark

Which of these is y^2-3y-10 factorised?

Hint

The two numbers must multiply to give -10 and add to give -3.

Worked solution

Which of these is y^2-3y-10 factorised?

  1. (y+5)(y-2)
  2. (y-5)(y-2)
  3. (y-5)(y+2)
  4. (y-10)(y+1)
  1. -5\times2=-10 and -5+2=-3, so y^2-3y-10=(y-5)(y+2).
  2. (y+5)(y-2) gives +3y, (y-5)(y-2) gives +10, and (y-10)(y+1) gives -9y.

Question 111 mark

x^2+2x-24=(x+6)(x-n)

Work out the value of n.

Hint

The two numbers in the brackets must multiply to give -24.

Worked solution

x^2+2x-24=(x+6)(x-n)

Work out the value of n.

  1. 6\times(-n)=-24, so n=4.
  2. Check: 6+(-4)=2, which matches the 2x.
  3. n=4

Question 123 marks

Leo says that

x^2+5x-14=(x-7)(x+2)

(a)

Which of these is true?

1 mark

(b)

Factorise x^2+5x-14 correctly.

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Work out what -7 and 2 multiply to give and add to give.

Worked solution

Leo says that

x^2+5x-14=(x-7)(x+2)

Part (a)

Which of these is true?

  1. Leo is right
  2. Leo is wrong: -7 and 2 add to give -5, not 5
  3. Leo is wrong: -7 and 2 multiply to give 14, not -14
  4. Leo is wrong: x^2 cannot be factorised
  1. -7\times2=-14, which is right.
  2. -7+2=-5, but the x term is +5x, so Leo is wrong.

Part (b)

Factorise x^2+5x-14 correctly.

  1. The numbers must multiply to give -14 and add to give 5.
  2. 7\times(-2)=-14 and 7+(-2)=5.
  3. So x^2+5x-14=(x+7)(x-2).

Question 132 marks

Factorise x^2-x-42

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

-x means -1x, so you need two numbers that multiply to give -42 and add to give -1.

Worked solution

Factorise x^2-x-42

  1. Pairs that multiply to give 42 include 6 and 7, which differ by 1.
  2. -7\times6=-42 and -7+6=-1.
  3. So x^2-x-42=(x-7)(x+6).

Question 142 marks

Here is a square.

The area of the square is (x^2+12x+36) cm².

Find an expression, in terms of x, for the length of one side of the square.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Factorise the area. The sides of a square are all the same length.

Worked solution

Here is a square.

The area of the square is (x^2+12x+36) cm².

Find an expression, in terms of x, for the length of one side of the square.

  1. Two numbers that multiply to give 36 and add to give 12 are 6 and 6.
  2. So x^2+12x+36=(x+6)(x+6).
  3. Area = side \times side, so each side is (x+6) cm.

Question 152 marks

x^2+bx+18 can be factorised as (x+p)(x+q), where p and q are positive whole numbers.

Which of these could not be the value of b?

Hint

List the pairs of positive whole numbers that multiply to give 18, and add each pair.

Worked solution

x^2+bx+18 can be factorised as (x+p)(x+q), where p and q are positive whole numbers.

Which of these could not be the value of b?

  1. 19
  2. 7
  3. 11
  4. 9
  1. p\times q=18 and p+q=b.
  2. 1\times18: 1+18=19. 2\times9: 2+9=11. 3\times6: 3+6=9.
  3. So b could be 19, 11 or 9, but not 7.

Question 16Challenge3 marks

Every term of 2x^2+14x+20 is a multiple of 2

(a)

Factorise 2x^2+14x+20 by taking out the common factor 2

1 mark

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(b)

Now factorise fully 2x^2+14x+20

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Divide every term by 2 to find what goes inside the bracket.

Worked solution

Every term of 2x^2+14x+20 is a multiple of 2

Part (a)

Factorise 2x^2+14x+20 by taking out the common factor 2

  1. 2x^2\div2=x^2, 14x\div2=7x and 20\div2=10.
  2. So 2x^2+14x+20=2(x^2+7x+10).

Part (b)

Now factorise fully 2x^2+14x+20

  1. Factorise the quadratic in the bracket: two numbers that multiply to give 10 and add to give 7 are 2 and 5.
  2. So x^2+7x+10=(x+2)(x+5).
  3. 2x^2+14x+20=2(x+2)(x+5).

Question 17Challenge3 marks

x^2-2x+k=(x-6)(x+m)

k and m are numbers.

(a)

Work out the value of m.

2 marks

(b)

Work out the value of k.

1 mark

Hint

The numbers -6 and m must add to give the number in front of x.

Worked solution

x^2-2x+k=(x-6)(x+m)

k and m are numbers.

Part (a)

Work out the value of m.

  1. Expanding, the x terms are mx-6x, so m-6=-2.
  2. So m=4.

Part (b)

Work out the value of k.

  1. k is the number on the end: -6\times m=-6\times4=-24.
  2. Check: (x-6)(x+4)=x^2+4x-6x-24=x^2-2x-24.
  3. k=-24

Question 18Challenge4 marks

Here is a rectangle.

The area of the rectangle is (x^2+11x+28) cm².

The shorter side has length (x+4) cm.

(a)

Find an expression, in terms of x, for the length of the longer side.

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(b)

Find an expression, in terms of x, for the perimeter of the rectangle. Give your answer in its simplest form.

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Factorise the area: one of the brackets will be (x+4).

Worked solution

Here is a rectangle.

The area of the rectangle is (x^2+11x+28) cm².

The shorter side has length (x+4) cm.

Part (a)

Find an expression, in terms of x, for the length of the longer side.

  1. Two numbers that multiply to give 28 and add to give 11 are 4 and 7.
  2. x^2+11x+28=(x+4)(x+7).
  3. So the longer side is (x+7) cm.

Part (b)

Find an expression, in terms of x, for the perimeter of the rectangle. Give your answer in its simplest form.

  1. Perimeter =(x+4)+(x+7)+(x+4)+(x+7).
  2. x+x+x+x=4x and 4+7+4+7=22.
  3. Perimeter =4x+22 cm.

Question 19Challenge3 marks

A square with sides of 11 cm is cut from the corner of a larger square with sides of x cm.

The shape that is left is shaded.

(a)

Find an expression, in terms of x, for the shaded area. Give your answer without brackets.

1 mark

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(b)

Factorise your answer to part (a).

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Area of the shaded shape = area of the large square - area of the small square.

Worked solution

A square with sides of 11 cm is cut from the corner of a larger square with sides of x cm.

The shape that is left is shaded.

Part (a)

Find an expression, in terms of x, for the shaded area. Give your answer without brackets.

  1. Large square: x\times x=x^2. Small square: 11\times11=121.
  2. Shaded area =x^2-121 cm².

Part (b)

Factorise your answer to part (a).

  1. The numbers must multiply to give -121 and add to give 0.
  2. 11\times(-11)=-121 and 11+(-11)=0.
  3. So x^2-121=(x+11)(x-11).

Question 20Challenge3 marks

Mia wants an expression that factorises with the same number in both brackets, as (x+a)(x+a)

(a)

Which of these expressions should Mia choose?

1 mark

(b)

Factorise the expression Mia should choose.

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

The two numbers in the brackets are the same, so they must add to give the number in front of x and multiply to give the number on the end.

Worked solution

Mia wants an expression that factorises with the same number in both brackets, as (x+a)(x+a)

Part (a)

Which of these expressions should Mia choose?

  1. x^2+14x+14
  2. x^2+14x+28
  3. x^2+14x+49
  4. x^2+49
  1. The two equal numbers add to give 14, so each one is 7.
  2. They must multiply to give 7\times7=49, so Mia should choose x^2+14x+49.
  3. 14 and 28 are not 7\times7, and x^2+49 has no x term, so its numbers would have to add to give 0.

Part (b)

Factorise the expression Mia should choose.

  1. Two numbers that multiply to give 49 and add to give 14 are 7 and 7.
  2. x^2+14x+49=(x+7)(x+7)=(x+7)^2.

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