Answers − Solving quadratics of the form ax² + bx + c = 0 by factorising first

1.   Solve 2x² + 10x + 8 = 0

i.		From answer 1 in Factorising Quadratics − 2 the factors are: (2x + 2)(x + 4)
ii.		So (2x + 2)(x + 4) = 0

iii.

Either

Verify by putting these two values of x into 2x² + 10x + 8

2.   Find the values of x where 3x² + 21x + 18 = 0

i.		From answer 2 in Factorising Quadratics − 2 the factors are: (3x + 3)(x + 6)
ii.		So (3x + 3)(x + 6) = 0

iii.

Either

Verify by putting these two values of x into 3x² + 21x + 18

3.   Solve 4x² + 20x − 24 = 0

i.		From answer 3 in Factorising Quadratics − 2 the factors are: (4x − 4)(x + 6)
ii.		So (4x − 4)(x + 6) = 0

iii.

Either

Verify by putting these two values of x into 4x² + 20x − 24

4.   Find the values of x where 2x² + 9x − 56 = 0

i.		From answer 4 in Factorising Quadratics − 2 the factors are: (2x − 7)(x + 8)
ii.		So (2x − 7)(x + 8) = 0

iii.

Either

Verify by putting these two values of x into 2x² + 9x − 56

5.   Solve 5x² − 20x − 60 = 0

i.		From answer 5 in Factorising Quadratics − 2 the factors are: (5x + 10)(x − 6)
ii.		So (5x + 10)(x − 6) = 0

iii.

Either

Check by putting these two values of x into 5x² − 20x − 60

6.   Find the values of x where 3x² − 21x + 36 = 0

i.		From answer 6 in Factorising Quadratics − 2 the factors are: (3x − 12)(x − 3)
ii.		So (3x − 12)(x − 3) = 0

iii.

Either

Check by putting these two values of x into 3x² − 21x + 36

7.   Solve 3x² − 40x + 48 = 0

i.		From answer 7 in Factorising Quadratics − 2 the factors are: (3x − 4)(x − 12)
ii.		So (3x − 4)(x − 12) = 0

iii.

Either

Check by putting these two values of x into 3x² − 40x + 48

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