Question

Factor Sin^2X + 7 Cos X + 7

Answer 1

`\begin{equation*}% \mbox{\large\( % - \left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-8\right)\)} %\end{equation*}`

Explanation:

sin²(x) = 1 - cos²(x)

So the expression sin²(x) + 7cosx +7 becomes

1 - cos²(x) + 7cos(x) + 7

Group like terms

`\begin{equation*}% \mbox{\large\( %=-\cos ^2\left(x\right)+7\cos \left(x\right)+1+7 %\)} %\end{equation*}`

`\begin{equation*}% \mbox{\large\( %=-\cos ^2\left(x\right)+7\cos \left(x\right)+8 %\)} %\end{equation*}`
`\begin{equation*}% \mbox{\large\( %=-\cos ^2\left(x\right)+7\cos \left(x\right)+1+7 %\)} %\end{equation*}`
`\begin{equation*}% \mbox{\large\( %=7\cos \left(x\right)-\cos ^2\left(x\right)+8\)} %\end{equation}`

`\begin{equation*}% \mbox{\large\( %=-\left(\cos ^2\left(x\right)-7\cos \left(x\right)-8\right)\)} %\end{equation}`

Let's deal with the expression in parenthesis

`\begin{equation}% \mbox{\large\( %=\cos ^2\left(x\right)-7\cos \left(x\right)-8\)} %\end{equation}`
`\begin{equation*}% \mbox{\large\( %=\cos ^2\left(x\right)-7\cos \left(x\right)-8)} %\end{equation}`

`\begin{equation*}% \mbox{\large\( %\cos^2(x) - 7\cos(x)-8\)} %\end{equation*}\\\\`

This is of the form
`\begin{equation*}% \mbox{\large\( %ax^2 + bx + c\)} %\end{equation*}`

with a = 1, b = -7 and c = -8

To factor this find two values u and v such that

`\begin{equation*}% \mbox{\large\( %u\cdot v = -8, v + v = -7\)} %\end{equation*}`

If we choose u = 1 and v = -8 we can satisfy the above relationship since
u · v = 1 (-8) = -8 and u + v = 1 +(-8) = -7

`\displaysize {\textrm{Group\:into\:}\left(ax^2+ux\right)+\left(vx+c\right)}`

`\begin{equation*}% \mbox{\large\( %\left(\cos ^2\left(x\right)+\cos \left(x\right)\right)+\left(-8\cos \left(x\right)-8\right)\)} %\end{equation*}`

`\textrm{Factor\:out}\:\cos \left(x\right)\:\mathrm{from}\:\cos ^2\left(x\right)+\cos \left(x\right)`

`\begin{equation*}% \mbox{\large\( %=\cos \left(x\right)\left(\cos \left(x\right)+1\right)\)} %\end{equation*}`

`\textrm{Factor\:out}\:-8\:\mathrm{from}\:-8\cos \left(x\right)-8:`

`=-8\left(\cos \left(x\right)+1\right)`

So

`\begin{equation*}% \mbox{\large\( %\left(\cos ^2\left(x\right)+\cos \left(x\right)\right)+\left(-8\cos \left(x\right)-8\right)\)} %\end{equation*}`

=
`\begin{equation*}% \mbox{\large\( %\cos \left(x\right)\left(\cos \left(x\right)+1\right)-8\left(\cos \left(x\right)+1\right)\)} %\end{equation*}`

`\textrm{Factor\:out\:common\:term\:}\cos \left(x\right)+1`

`\begin{equation*}% \mbox{\large\( %=\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-8\right)\)} %\end{equation*}`

Since this was the term in the parenthesis with a leading negative sign, add the negative sign to get

`\begin{equation*}% \mbox{\large\( %1-\cos ^2\left(x\right)+7\cos \left(x\right)+7\)} %\end{equation*}`
`\begin{equation*}% \mbox{\large\( %= - \left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-8\right)\)} %\end{equation*}`
`\begin{equation*}% \mbox{\large\( %= - \left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-8\right)\)} %\end{equation*}\\`

Answer:
`\begin{equation*}% \mbox{\large\( % - \left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-8\right)\)} %\end{equation*}`