Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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.

share|cite|improve this question
up vote 17 down vote accepted

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





share|cite|improve this answer

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

share|cite|improve this answer

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 $t$:$$ (a^2 + 2a - 3)(a^2 + 2a+1)$$


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

share|cite|improve this answer

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.

share|cite|improve this answer
No need to guess: the rational root theorem says every rational root of $f$ is on an explicit list of 4 possibilities. – Hurkyl Mar 27 '13 at 8:28

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.