National 5 Maths
Expanding Brackets

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

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

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

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

We can use the FOIL method: Firsts, Outsides, Insides, Lasts.

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

Alternatively, we can use the distributive method:

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

Again, we can use the FOIL method: Firsts, Outsides, Insides, Lasts.

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

Alternatively, we can use the distributive method:

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

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

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

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

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

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

Other methods exist! See examples 5 and 6 above.

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

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

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

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

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

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