National 5 Maths
Expanding Brackets

This type of expansion is really just two 'common factor' type expansions, one after the other.

So the \(5\) is multiplied by each term in the first bracket and the \(+3\) is multiplied by each term in the second bracket.

$$ \begin{eqnarray} &\ & \overparen{\overparen{5(x}+4})+\overparen{\overparen{3(2x}+1})\\[8pt] &=& 5x+20+6x+3\\[8pt] &=& \:11x+23 \end{eqnarray} $$

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This is the same type as the previous example, but we need to handle the negative numbers correctly.

$$ \begin{eqnarray} &\ &\overparen{\overparen{-2(3x}+4})+\overparen{\overparen{5(2x}-5})\\[8pt] &=&-\!6x-8+10x-25\\[8pt] &=&\:4x-33 \end{eqnarray} $$

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This type is very simple. We just multiply the \(x\) that is outside the bracket by each term that is inside.

$$ \begin{eqnarray} &\ & \overparen{\overparen{x(3x}+4})\\[8pt] &=&\ 3x^2+4x \end{eqnarray} $$

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This is the same type as the previous example, but it we need to multiply by \(6\) as well as \(x\small,\) and we must handle the negative coefficient correctly.

$$ \begin{eqnarray} &\ &\overparen{\overparen{6x(-2x}+3})\\[8pt] &=&\ -\!12x^2+18x \end{eqnarray} $$

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There are three main methods for this type of 'double bracket' expansion with two terms in each bracket.

1. The grid method

This method may be best for students who like to visualise the fact that each term in the first bracket is multiplied by each term in the second bracket.

$$ \begin{array} {|c|c|} \hline & \textcolor{red}{x} & \textcolor{red}{+2} \\ \hline \textcolor{red}{x} & x^2 & +2x \\ \hline \textcolor{red}{+5} & +5x & +10 \\ \hline \end{array} $$

Combining the \(+2x\) and \(+5x\) terms, we obtain \(x^2+7x+10\small.\)

2. The distributive method

This is also a good method. We split the terms of the first bracket and multiply each of them by the terms in the second bracket. This turns the expression into the same form as worked examples 1 and 2 above.

$$ \begin{eqnarray} &\ & (x+2)(x+5)\\[8pt] &=&\ x(x+5) + 2(x+5)\\[8pt] &=&\ x^2+5x+2x+10\\[8pt] &=&\ x^2+7x+10 \end{eqnarray} $$

3. The 'FOIL' method

FOIL stands for 'Firsts, Outsides, Insides, Lasts'. It is a quick way of multiplying each term in the first bracket by each term in the second.

Some teachers think that FOIL is a terrible method, because it won't extend to more than two terms in either bracket. We're fine with it, but you should also understand the other two methods. FOIL is a handy shortcut for the specific, and very common, situation in which there are two terms in each bracket.

$$ \begin{eqnarray} &\ & (x+2)(x+5)\\[8pt] &=&\ \underbrace{x^2}_{\textsf{F}}\:\:\underbrace{+\ 5x}_{\textsf{O}}\:\:\underbrace{+\ 2x}_{\textsf{I}}\:\:\underbrace{+\ 10}_{\textsf{L}}\\[8pt] &=&\ x^2\:\underbrace{+\ 7x}_{\textsf{OI}}\:+\ 10 \end{eqnarray} $$

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As for the previous example, there are three main methods of expanding the expression.

1. The grid method

$$ \begin{array} {|c|c|} \hline & \textcolor{red}{-x} & \textcolor{red}{+6} \\ \hline \textcolor{red}{3x} & -3x^2 & +18x \\ \hline \textcolor{red}{-5} & +5x & -30 \\ \hline \end{array} $$

Combining the \(+18x\) and \(+5x\small,\) we obtain \(-3x^2+23x-30\small.\)

2. The distributive method

$$ \begin{eqnarray} &\ & (-x+6)(3x-5)\\[8pt] &=&\ -\!x(3x-5) + 6(3x-5)\\[8pt] &=&\ -\!3x^2+5x+18x-30\\[8pt] &=&\ -\!3x^2+23x-30 \end{eqnarray} $$

3. The 'FOIL' method

$$ \begin{eqnarray} &\ & (-x+6)(3x-5)\\[8pt] &=&\ \underbrace{-3x^2}_{\textsf{F}}\:\:\underbrace{+\ 5x}_{\textsf{O}}\:\:\underbrace{+\ 18x}_{\textsf{I}}\:\:\underbrace{-\ 30}_{\textsf{L}}\\[8pt] &=&\ -\!3x^2\:\underbrace{+\ 23x}_{\textsf{OI}}\:-\ 30 \end{eqnarray} $$

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The grid method can be used for this question, as for all expansion questions. Go ahead, if you prefer!

Note that we cannot use 'FOIL' for this type. It can only be used when there are two terms in each bracket.

The distributive method might be quickest. Split the first bracket into its two terms and then multiply each term by the second bracket.

$$ \begin{eqnarray} &\ & (x+2)(x^2+x+3)\\[8pt] &=&\ \overparen{\overparen{\overparen{x(x^2}+x}+3})\ \overparen{\overparen{\overparen{+2(x^2}+x}+3})\\[8pt] &=&\ x^3+x^2+3x+2x^2+2x+6\\[8pt] &=&\ x^3+3x^2+5x+6 \end{eqnarray} $$

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Again, we can split the first bracket into its two terms and then multiply each term by the second bracket.

$$ \begin{eqnarray} &\ & (2x-1)(7x^2+x-5)\\[8pt] &=&\ \overparen{\overparen{\overparen{2x(7x^2}+x}-5})\ \overparen{\overparen{\overparen{-1(7x^2}+x}-5})\\[8pt] &=&\ 14x^3+2x^2-10x-7x^2-x+5\\[8pt] &=&\ 14x^3-5x^2-11x+5 \end{eqnarray} $$

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$$ \begin{eqnarray} &\ & (2x-5)(3x+1)\\[6pt] &=&\ 6x^2+2x-15x-5\\[6pt] &=&\ 6x^2-13x-5 \end{eqnarray} $$

Other methods exist! See examples 5 and 6 above.

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$$ \begin{eqnarray} &\ & (x-4)(x^2+x-2)\\[6pt] &=&\ x(x^2+x-2)-4(x^2+x-2)\\[6pt] &=&\ x^3+x^2-2x-4x^2-4x+8\\[6pt] &=&\ x^3-3x^2-6x+8 \end{eqnarray} $$

You could also use the grid/box method for this question, if you understand it better.

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$$ \begin{eqnarray} &\ & (2x+3)(x^2-4x+1)\\[6pt] &=&\ 2x(x^2-4x+1)+3(x^2-4x+1)\\[6pt] &=&\ 2x^3-8x^2+2x+3x^2-12x+3\\[6pt] &=&\ 2x^3-5x^2-10x+3 \end{eqnarray} $$

You could also use the grid/box method for this question, if you understand it better.

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$$ \begin{eqnarray} &\ & (3x+1)(x-1)+2(x^2-5)\\[6pt] &=&\ 3x^2-3x+x-1+2x^2-10\\[6pt] &=&\ 5x^2-2x-11 \end{eqnarray} $$

You could also use two grids/boxes for this question, if you understand that better, but you will still need to collect all of the like terms and give your final answer in the fully simplified form.

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$$ \begin{eqnarray} &\ & (x+5)(2x^2-7x-3)\\[6pt] &=&\ x(2x^2-7x-3)+5(2x^2-7x-3)\\[6pt] &=&\ 2x^3-7x^2-3x+10x^2-35x-15\\[6pt] &=&\ 2x^3+3x^2-38x-15 \end{eqnarray} $$

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$$ \begin{eqnarray} &\ & (x+7)^2+6(x^2-10)\\[6pt] &=&\ (x+7)(x+7)+6x^2-60\\[6pt] &=&\ x^2+7x+7x+49+6x^2-60\\[6pt] &=&\ 7x^2+14x-11\\[6pt] \end{eqnarray} $$

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$$ \begin{eqnarray} &\ & (x+1)(x^2-4x+5)\\[6pt] &=&\ x(x^2-4x+5)+1(x^2-4x+5)\\[6pt] &=&\ x^3-4x^2+5x+x^2-4x+5\\[6pt] &=&\ x^3-3x^2+x+5 \end{eqnarray} $$

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$$ \begin{eqnarray} &\ & (x+3)(x+5)+4(x-2)\\[6pt] &=&\ x^2+5x+3x+15+4x-8\\[6pt] &=&\ x^2+12x+7 \end{eqnarray} $$

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$$ \begin{eqnarray} &\ & (y+4)(y^2-3y+2)\\[6pt] &=&\ y(y^2-3y+2)+4(y^2-3y+2)\\[6pt] &=&\ y^3-3y^2+2y+4y^2-12y+8\\[6pt] &=&\ y^3+y^2-10y+8 \end{eqnarray} $$

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