# Express x^{2}+8x in the form (x+a)2+b

Express $$\displaystyle{x}^{{{2}}}+{8}{x}$$  in the form  $$\displaystyle{\left({x}+{a}\right)}{2}+{b}$$

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hesgidiauE

To express $$\displaystyle{x}^{{{2}}}+{8}{x}$$ in the form $$(x+a)^{2}+b,$$ we need to complete the square.
For an expression of the form $$\displaystyle{x}^{{{2}}}+{B}{x}$$, we can complete the square by adding and subtracting $$\displaystyle{\frac{{{B}}}{{{2}}}}^{{{2}}}$$. You need to add and subtract this number so the new expression will be equivalent to the original expression.
For $$x^{2}+8x, B=8$$ so we need to add and subtract $$(82)2=42=16$$. This gives:
$$\displaystyle{x}^{{{2}}}+{8}{x}={x}^{{{2}}}+{8}{x}+{16}−{16}$$
Group the first three terms to form a perfect square trinomial:
$$\displaystyle={\left({x}^{{2}}+{8}{x}+{16}\right)}−{16}$$
A perfect square trinomial of the form $$\displaystyle{x}^{{{2}}}+{B}{x}+{\frac{{{B}}}{{{2}}}}^{{{2}}}$$ factors to $$\displaystyle{\left({x}+{B}{2}\right)}{2}{\left({x}+{2}{B}\right)}^{{{2}}}$$.
Since $$\displaystyle{\frac{{{B}}}{{{2}}}}={\frac{{{8}}}{{{2}}}}={4},{t}{h}{e}{n}\ {x}^{{{2}}}+{8}{x}+{16}$$ factors to $$\displaystyle{\left({x}+{4}\right)}^{{{2}}}$$. Therefore:
$$\displaystyle{\left({x}^{{{2}}}+{8}{x}+{16}\right)}−{16}={\left({x}+{4}\right)}^{{{2}}}−{16}$$
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