How do you factor ${cos}^{2} x + 7 cos x + 8$ $cos^2 x+7 cos x+8$ ?

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How do you factor ${cos}^{2} x + 7 cos x + 8$ $cos^2 x+7 cos x+8$ ?

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Answer 1 · 2018-05-30T01:34:33

$\frac{1}{4} (2 cos x + 7 + \sqrt{17}) (2 cos x + 7 - \sqrt{17})$ $1/4(2cosx+7+sqrt17)(2cosx+7-sqrt17)$

Explanation:

First let $t = cos x$ $t=cosx$ .

$y = t^{2} + 7 t + 8$ $y=t^2+7t+8$

Now, let's complete the square to factor this.

$y = (t^{2} + 7 t) + 8$ $y=(t^2+7t)+8$

Note that ${(t + \frac{7}{2})}^{2} = (t + \frac{7}{2}) (t + \frac{7}{2})$ $(t+7/2)^2=(t+7/2)(t+7/2)$

$= t^{2} + \frac{7}{2} t + \frac{7}{2} t + {(\frac{7}{2})}^{2}$ $=t^2+7/2t+7/2t+(7/2)^2$

$= t^{2} + 7 t + \frac{49}{4}$ $=t^2+7t+49/4$

So we want to add $\frac{49}{4}$ $49/4$ into the expression and subtract it back out again.

$y = (t^{2} + 7 t + \frac{49}{4}) + 8 - \frac{49}{4}$ $y=(t^2+7t+49/4)+8-49/4$

Note that $8 - \frac{49}{4} = \frac{32}{4} - \frac{49}{4} = - \frac{17}{4}$ $8-49/4=32/4-49/4=-17/4$ .

$y = {(t + \frac{7}{2})}^{2} - \frac{17}{4}$ $y=(t+7/2)^2-17/4$

Now, note that $\frac{17}{4} = {(\frac{\sqrt{17}}{2})}^{2}$ $17/4=(sqrt17/2)^2$ .

$y = {(t + \frac{7}{2})}^{2} - {(\frac{\sqrt{17}}{2})}^{2}$ $y=(t+7/2)^2-(sqrt17/2)^2$

Now, we have a difference of squares and can factor it as one.

$y = [(t + \frac{7}{2}) + \frac{\sqrt{17}}{2}] [(t + \frac{7}{2}) - \frac{\sqrt{17}}{2}]$ $y=[(t+7/2)+sqrt17/2][(t+7/2)-sqrt17/2]$

$y = (cos x + \frac{7 + \sqrt{17}}{2}) (cos x + \frac{7 - \sqrt{17}}{2})$ $y=(cosx+(7+sqrt17)/2)(cosx+(7-sqrt17)/2)$

If we wish, we can bring a common factor of $\frac{1}{2}$ $1/2$ out of each part:

$y = \frac{1}{4} (2 cos x + 7 + \sqrt{17}) (2 cos x + 7 - \sqrt{17})$ $y=1/4(2cosx+7+sqrt17)(2cosx+7-sqrt17)$

Answer 2 · 2018-05-30T02:00:37

$(cos (x) + \frac{7 + \sqrt{17}}{2}) (cos (x) + \frac{7 - \sqrt{17}}{2})$ $(cos(x) + \frac{7 + \sqrt(17)}{2})(cos(x) + \frac{7 - \sqrt(17)}{2})$

Explanation:

let $u = cos (x)$ $u = cos(x)$
The question then becomes:

Factor $u^{2} + 7 u + 8$ $u^2+7u+8$ you could just use quadratic formula here i.e. $u = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $u = \frac{-b\pm \sqrt(b^2-4ac)}{2a}$

or you could do it the long way (which isn't any better than the formula, in fact it's one of the methods use to formulate the quadratic formula) :
find two roots, $r_{1}$ $r_1$ and $r_{2}$ $r_2$ such that $(u - r_{1}) (u - r_{2}) = u^{2} + 7 u + 8$ $(u-r_1)(u - r_2) = u^2+7u+8$

Expand: $(u - r_{1}) (u - r_{2}) = u^{2} - r_{1} u - r_{2} u + (r_{1}) (r_{2})$ $(u-r_1)(u - r_2) = u^2 - r_1u - r_2u + (r_1)(r_2)$
$= u^{2} - (r_{1} + r_{2}) u + (r_{1}) (r_{2})$ $= u^2 - (r_1+r_2)u + (r_1)(r_2)$

Thus: $u^{2} - (r_{1} + r_{2}) u + (r_{1}) (r_{2}) = u^{2} + 7 u + 8$ $u^2 - (r_1+r_2)u + (r_1)(r_2) = u^2+7u+8$
and therefore: $- (r_{1} + r_{2}) = 7$ $- (r_1+r_2) = 7$ and $(r_{1}) (r_{2}) = 8$ $(r_1)(r_2) = 8$

$(r_{1} + r_{2}) = - 7, {(r_{1} + r_{2})}_{1}^{2} = 49$ $(r_1+r_2) = -7, (r_1+r_2)^2 = 49$
${(r_{1})}_{1}^{2} + 2 (r_{1}) (r_{2}) + {(r_{2})}_{2}^{2} = 49$ $(r_1)^2 + 2(r_1)(r_2) + (r_2)^2 = 49$
${(r_{1})}_{1}^{2} + 2 (r_{1}) (r_{2}) + {(r_{2})}_{2}^{2} - 4 (r_{1}) (r_{2}) = 49 - 4 (8) = 17$ $(r_1)^2 + 2(r_1)(r_2) + (r_2)^2 - 4(r_1)(r_2) = 49 - 4(8) = 17$
${(r_{1})}_{1}^{2} - 2 (r_{1}) (r_{2}) + {(r_{2})}_{2}^{2} = 17$ $(r_1)^2 - 2(r_1)(r_2) + (r_2)^2 = 17$

$(r_1-r_2)^2 = 17$
$r_1-r_2 = \sqrt(17)$
$\frac{r_1+r_2 + r_1-r_2}{2} = r_1 = \frac{-7 + \sqrt(17)}{2}$
$\frac{r_1+r_2 - (r_1-r_2)}{2} = r_2 = \frac{-7 - \sqrt(17)}{2}$

Thus, the factored form is $(u + \frac{7 + \sqrt(17)}{2})(u + \frac{7 - \sqrt(17)}{2})$

sub $u = cos(x)$ to get:

$(cos(x) + \frac{7 + \sqrt(17)}{2})(cos(x) + \frac{7 - \sqrt(17)}{2})$

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How do you factor ${cos}^{2} x + 7 cos x + 8$ $cos^2 x+7 cos x+8$ ?

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