Factorising x² + bx + c

\( x^2+bx+c=(x+p)(x+q),\;p+q=b,\;pq=c \)
Algebra GCSE
Question 1 of 20

\( Factorise x^2-10x+21 \)

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
Numbers multiply to 21 and add to -10.

Explanation

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Statement

A quadratic expression of the form:

\[ x^2 + bx + c \]

can be factorised into two brackets:

\[ x^2 + bx + c = (x+p)(x+q) \]

where the numbers \(p\) and \(q\) satisfy the conditions:

  • \(p + q = b\)
  • \(p \times q = c\)

Why it works

  • Expanding \((x+p)(x+q)\) gives \(x^2 + (p+q)x + pq\).
  • This matches the original quadratic \(x^2 + bx + c\) when \(p+q = b\) and \(pq = c\).
  • Therefore, finding the correct pair of numbers \(p, q\) allows us to factorise the quadratic into brackets.

Recipe (Steps to Factorise)

  1. Write down the quadratic \(x^2 + bx + c\).
  2. Find two numbers \(p, q\) such that:
    • \(p+q = b\)
    • \(p \times q = c\)
  3. Write the factorised form as \((x+p)(x+q)\).
  4. Check by expanding to confirm it matches the original quadratic.

Spotting it

Use this method when the coefficient of \(x^2\) is 1 (monic quadratics). If the coefficient is not 1, you need other methods such as grouping or the quadratic formula.

Common pairings

  • Solving quadratic equations (\(x^2 + bx + c = 0\)).
  • Sketching quadratic graphs (roots come from the factors).
  • Simplifying algebraic fractions.

Mini examples

  1. \(x^2 + 5x + 6 = (x+2)(x+3)\), since 2+3=5 and 2×3=6.
  2. \(x^2 - x - 6 = (x-3)(x+2)\), since -3+2=-1 and -3×2=-6.

Pitfalls

  • Forgetting that signs matter (both addition and multiplication conditions must hold).
  • Mixing up the order — although \((x+p)(x+q)\) = \((x+q)(x+p)\).
  • Trying to factorise quadratics with no integer factors (these need quadratic formula or completing the square).

Exam Strategy

  • Write down factor pairs of \(c\) to test possible values quickly.
  • Check sum = \(b\), product = \(c\).
  • Expand your final brackets to verify correctness.

Worked Micro Example

Factorise \(x^2 + 7x + 10\).

We need two numbers that multiply to 10 and add to 7: \(5\) and \(2\).

So, \(x^2 + 7x + 10 = (x+5)(x+2)\).

Summary

To factorise \(x^2 + bx + c\), find two numbers whose product is \(c\) and sum is \(b\). This quick method works for monic quadratics and is essential for solving quadratic equations and simplifying algebraic expressions.

Worked examples

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  1. \( Factorise x^2+5x+6 \)
    1. \( Find numbers with sum=5, product=6 \)
    2. \( 2+3=5,2*3=6 \)
    3. (x+2)(x+3)
    Answer: (x+2)(x+3)
  2. \( Factorise x^2+7x+10 \)
    1. \( Sum=7,product=10 \)
    2. \( 5+2=7,5*2=10 \)
    3. (x+5)(x+2)
    Answer: (x+5)(x+2)
  3. \( Factorise x^2+8x+15 \)
    1. \( Sum=8,product=15 \)
    2. \( 3+5=8,3*5=15 \)
    3. (x+3)(x+5)
    Answer: (x+3)(x+5)
  4. \( Factorise x^2+6x+9 \)
    1. \( Sum=6,product=9 \)
    2. \( 3+3=6,3*3=9 \)
    3. (x+3)(x+3)
    Answer: \( (x+3)^2 \)
  5. \( Factorise x^2-5x+6 \)
    1. \( Sum=-5,product=6 \)
    2. \( -2+-3=-5,-2*-3=6 \)
    3. (x-2)(x-3)
    Answer: (x-2)(x-3)
  6. \( Factorise x^2-x-6 \)
    1. \( Sum=-1,product=-6 \)
    2. \( -3+2=-1,-3*2=-6 \)
    3. (x-3)(x+2)
    Answer: (x-3)(x+2)
  7. \( Factorise x^2+4x-12 \)
    1. \( Sum=4,product=-12 \)
    2. \( 6+(-2)=4,6*-2=-12 \)
    3. (x+6)(x-2)
    Answer: (x+6)(x-2)
  8. \( Factorise x^2-7x+12 \)
    1. \( Sum=-7,product=12 \)
    2. \( -3+-4=-7,-3*-4=12 \)
    3. (x-3)(x-4)
    Answer: (x-3)(x-4)
  9. \( Factorise x^2-3x-28 \)
    1. \( Sum=-3,product=-28 \)
    2. \( 4+(-7)=-3,4*-7=-28 \)
    3. (x+4)(x-7)
    Answer: (x+4)(x-7)
  10. \( Factorise x^2+2x-15 \)
    1. \( Sum=2,product=-15 \)
    2. \( 5+(-3)=2,5*-3=-15 \)
    3. (x+5)(x-3)
    Answer: (x+5)(x-3)