$$(a^{ 2 }+1)^{ 2 }-7(a^{ 2 }+1)+10$$
So far I got:
$$(a^{ 2 }+1)(a^2+1)-7a^{ 2 }+3$$
I feel like I am going about this the wrong way. I need a push in the right direction.
3 Answers
(STRONG) HINT: You have a quadratic $x^2-7x+10$ in the variable $x=a^2+1$. Find x first using the quadratic formula and then solve for a using the equation $x=a^2+1$. This will give you the factors.
$$(a^{ 2 }+1)^{ 2 }-7(a^{ 2 }+1)+10$$
put x=$a^{2}$,then we get
$$\implies \ (x+1)^{2}-7(x+1)+10$$ Now solve this
$$\implies\ x^2+2x+1-7x-7+10$$ $$\implies\ x^2-5x+4$$ $$\implies \ (x-4)(x-1)$$
i.e $$\implies\ (a^{2}-4)(a^{2}-1)$$ therefore four roots of the equation are: $$a=+2,-2,+1,-1$$
$$(a^2+1)^2-7(a^2+1)+10$$ $x=a^2+1$ $$x^2-7x+10=x^2-5x-2x+10=x(x-5)-2(x-5)=(x-2)(x-5)=$$ then $$(a^2+1)^2-7(a^2+1)+10=(a^2+1-2)(a^2+1-5)=(a^2-1)(a^2-4)=$$ $$=(a-1)(a+1)(a+2)(a-2)$$
