Expand the first two brackets by multiplying everything in the first bracket by everything in the second bracket.

$(x \times x + 5 \times x + x \times 2 + 5 \times 2) (x + a)$

Simplify by completing the multiplications.

$(x^{2} + 5 x + 2 x + 10) (x + a)$

Simplify the terms inside the first bracket by collecting "like" terms.

$(x^{2} + 7 x + 10) (x + a)$

Expand the two remaining brackets by multiplying all three terms inside the first bracket by each of the terms inside the second bracket.

$x^{2} \times x + x^{2} \times a + 7 x \times x + 7 x \times a + 10 \times x + 10 \times a$

Simplify by completing the multiplications.

$x^{3} + a x^{2} + 7 x^{2} + 7 a x + 10 x + 10 a$

Simplify by collecting "like terms".

$x^{3} + (a + 7) x^{2} + (7 a + 10) x + 10 a$

Equate the expanded expression to the given result.

$x^{3} + (a + 7) x^{2} + (7 a + 10) x + 10 a \equiv x^{3} + b x^{2} + c x - 30$

Compare the constant term on both sides.

$10 a = - 30$

Solve to find the values of $a$ .

$a = \frac{- 30}{10}$

$a = - 3$ [1]

Compare the $x^{2}$ term on both sides.

$\begin{array}{rcl} (a + 7) x^{2} & = & b x^{2} \\ a + 7 & = & b \end{array}$

Substitute $a = - 3$ into the equation and solve to find the value of $b$ .

$(- 3) + 7 = b$

$b = 4$ [1]

Compare the $x$ term on both sides.

$\begin{array}{rcl} (7 a + 10) x & = & c x \\ 7 a + 10 & = & c \end{array}$

Substitute $a = - 3$ into the equation and solve to find the value of $c$ .

$\begin{array}{rcl} 7 \times (- 3) + 10 & = & c \\ - 21 + 10 & = & c \end{array}$

$c = - 11$ [1]

Expanding Brackets (AQA GCSE Maths)

Exam Questions