How do you factor ${(a^{2} + 1)}^{2} - 7 (a^{2} + 1) + 10$ $(a^2 +1)^2 - 7(a^2 +1) +10$ ?

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How do you factor ${(a^{2} + 1)}^{2} - 7 (a^{2} + 1) + 10$ $(a^2 +1)^2 - 7(a^2 +1) +10$ ?

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Answer 1 · 2015-09-28T20:15:48

$a_{1} = 1, a_{2} = - 1, a_{3} = 2, a_{4} = - 2$ $a_1=1 , a_2=-1, a_3=2, a_4=-2$

Explanation:

Let $(a^{2} + 1) = x$ $(a^2+1) =x$
so the eqn is $x^{2} - 7 x + 10 = 0$ $x^2-7x+10=0$
now,
$x^{2} - 5 x - 2 x + 10 = 0$ $x^2 -5x-2x+10=0$
$\Rightarrow x (x - 5) - 2 (x - 5) = 0$ $=>x(x-5)-2(x-5)=0$
$\Rightarrow (x - 2) \cdot (x - 5) = 0$ $=>(x-2)*(x-5)=0$
$\Rightarrow x - 2 = 0 \Rightarrow x = 2 \Rightarrow a^{2} + 1 = 2 \Rightarrow a^{2} = 1$ $=>x-2=0 =>x=2 =>a^2+1=2 => a^2=1$
$\Rightarrow a = \pm 1$ $=>a=+-1$
Again,
$\Rightarrow x - 5 = 0 \Rightarrow x = 5 \Rightarrow a^{2} + 1 = 5 \Rightarrow a^{2} = 4$ $=>x-5=0 =>x=5 =>a^2+1=5 => a^2=4$
$\Rightarrow a = \pm 2$ $=>a=+-2$

Answer 2 · 2015-09-28T20:24:18

${(a^{2} + 1)}^{2} - 7 (a^{2} + 1) + 10 = (a + 1) (a - 1) (a + 2) (a - 2)$ $(a^2+1)^2-7(a^2+1)+10 = (a+1)(a-1)(a+2)(a-2)$

Explanation:

Let $u = a^{2} + 1$ $u = a^2+1$ , then the expression is:

$u^{2} - 7 u + 10$ $u^2-7u+10$ which can be factored:

$(u - 2) (u - 5)$ $(u-2)(u-5)$

Replacing $u$ $u$ , we get:

$((a^{2} + 1) - 2) ((a^{2} + 1) - 5)$ $((a^2+1)-2)((a^2+1)-5)$ .

We can simplify to get:

$(a^{2} - 1) (a^{2} - 4)$ $(a^2-1)(a^2-4)$ .

Each of these is a difference of squares, so we can finish with:

$(a + 1) (a - 1) (a + 2) (a - 2)$ $(a+1)(a-1)(a+2)(a-2)$

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How do you factor ${(a^{2} + 1)}^{2} - 7 (a^{2} + 1) + 10$ $(a^2 +1)^2 - 7(a^2 +1) +10$ ?

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How do you factor (a2+1)2−7(a2+1)+10(a^2 +1)^2 - 7(a^2 +1) +10?

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How do you factor ${(a^{2} + 1)}^{2} - 7 (a^{2} + 1) + 10$ $(a^2 +1)^2 - 7(a^2 +1) +10$ ?