Factor $(a^2+2a)^2-2(a^2+2a)-3$ completely

I have this question that asks to factor this expression completely:

$$(a^2+2a)^2-2(a^2+2a)-3$$

My working out:

$$a^4+4a^3+4a^2-2a^2-4a-3$$ $$=a^4+4a^3+2a^2-4a-3$$ $$=a^2(a^2+4a-2)-4a-3$$

I am stuck here. I don't how to proceed correctly.

Solution 1:

Although your working is correct, you won't be able to proceed further. The trick is to let $y=a^2+2a$. Then your expression becomes

$y^2-2y-3$

$=(y-3)(y+1)$

$=(a^2+2a-3)(a^2+2a+1)$

$=(a+3)(a-1)(a+1)^2$

Solution 2:

If you assign $$ a^2 + 2a = x $$ you'll get: $$ x^2 - 2x - 3 $$ Considering that $$ x^2 - 2x - 3 = (x - 3)(x+1) $$ you'll get: $$ (a^2 + 2a - 3)(a^2 + 2a+1) = (a + 3)(a - 1)(a + 1)^2 $$

Solution 3:

When you do such problems, first try to find a common term out. Here you can see that the common term couldn't be separated the easier way. So, the best way is to just take the most possible common term to a new variable.

Let the variable be $t$ here.

So, $$ t=a^2+2a$$ So, your equation becomes:

$$ t^2 - 2t - 3 $$ Now that's easy, you see. The equation is reduced to a quadratic equation. Now it can be easily solved.

$$ t^2-2t-3 = (t-3)(t+1)$$

At last plug in the value of $t$:$$ (a^2 + 2a - 3)(a^2 + 2a+1)$$

$$\implies(a+1)^2(a+3)(a-1)$$

P.S. In the meantime, I was writing the solution, Jasper and Hayk have also given the same solution.

Solution 4:

Or if you missed Jasper Loy's trick, you can guess and check a value of $a$ for which $$f(a) = a^4 +4a^3 +2a^2 −4a−3 = 0.$$

E.g. f(1) = 0 so $(a-1)$ is a factor and you can use long division to factorise it out.

Solution 5:

Often, a problem is handed to us in a slightly convenient form. Here, we may note the quadratic form: $$(a^2+2a)^2−2(a^2+2a)−3$$ We can guess it will factor into four factors, so let's find the four roots, via setting the equation to zero and solving. $$(a^2+2a)^2−2(a^2+2a)−3=0$$ Lets complete the square: $[(a^2+2a)-1]^2=4$, then $a^2+2a-1=\pm2$. We solve the two equations, $a^2+2a-3=0$ and $a^2+2a+1=0$. This will give real roots, so we can completely factor the above polynomial. I'll let you finish.