Factorise $x^4 - 3x^3 - 41x - 45 = 0$

Question

The answer is $(x^2 - 4x - 5)(x^2 + x + 9)$ but I don't know how to get there.

As a solution I tried this. Write the polynomial as:

$(x^2 + Ax + B)(x^2 + Cx + D) = 0$

Then equate powers of $x$ and solve the following system:

$x^3: A + C = -3$

$x^2: AC + D + B = 0$

$x^1: AD + BC = -41$

$x^0: BD = -45$

As far as I can see, this system of linear equations, produces a more complicated polynomial.

Answer 1

By the rational root theorem we find the roots $5$ and $-1$ so that the polynomial factors as $$ (x^2 + x + 9)(x + 1)(x - 5). $$ The only confusion here with our answer is that $x^2 - 4x - 5$ is not irreducible but still decomposes into two linear factors. Dividing them out you don't have to make the computation $(x^2+ax+b)(x^2+cx+d)$ any longer.

Answer 2

HINT

If the idea is to find integer factors, note that the free term $45$, must be the product of the individual roots, so you can get a clue by factoring it. Note $45 = 5\cdot 9 = 5 \cdot 3^2$, so it makes sense to try plugging in $\pm 1, \pm 3, \pm 5, \pm 9, \pm 15$.

If when you plug in a value $v$, you do get zero, the polynomial should factor by $x-v$.

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UPDATE

So you end up with the system $$ \begin{split} A + C &= -3\\ AC + D + B &= 0\\ AD + BC &= -41\\ BD &= -45 \end{split} $$ You can solve it by elimination. For example, from the first equation, you have $A = -3-C$ and from the last equation, you have $B = -45/D$. Plugging those into the other two equations will yield two equations in two unknowns...

Answer 3

$A+C = -3$ means $A = -(C+3)$

$BD = -45$ means $B=-\frac {45}D$

So

$AC + D+B =0$ means

$-(C+3)C + D -\frac {45}D =0$ means

$C^2 + 3C +(\frac {45}D-D)$ or that

$C = \frac {-3\pm \sqrt{9 + 4(D-\frac{45}D)}2$

Now we presumed that this could be factored so $D$ is an integer that divides $45$ and $9 + 4(D-\frac {45}D) \ge 0$.

If $|D|= 1$ then $0 \le 9+4(D-\frac {45}D)= 9+4(45-1)=185$ which is not a perfect square.

If $|D|=3$ then $0 \le 9+4(D-\frac {45}D) =9+4(15-3)=57$ which is not a perfect square.

If $|D|=5$ then $0 \le 9+ 4(9-5)=25= 5^2$.

So $D=-5$.

By symmetry, we know $D=\frac{45}{-5} = 9$ will yield same result and $|D| = 15, 45$ the same non results.

So $D = -5$ or $9$ and $B = 9$ or $-5$

And $C =\frac {-3 \pm \sqrt {25}}2= -4, or $1$.

Wolog we will do $D=-5$ and $B=9$.

The fourth equation is

$AD + BC =-41$ which give us

$-(C+3)(-5) + 9C = -41$

So $14C = -56$ or $C = -4$.

And that gives us $A=-(-4+3)=1$.

So $x^4 - 3x^3 - 41x -45 = (x^2 + x +9)(x -4x -5)$

.......

But for heck sake, why didn't you just use the rational root theorem?

If $x = -1$ then $x^4 - 3x^3 -41x -45=1+3+41-45=0$ so $(x+1)$ is a factor so $x^4-3x^3 -41x -45 = x^3(x+1) - 4x^3 - 41x -45=$

$x^3(x+1)- 4x^2(x+1) + 4x^2-41x -45=$

$x^3(x+1)-4x^2(x+1) + 4x(x+1) -45x-45=$

$x^3(x+1) - 4x^2(x+1) + 4x(x+1) - 45(x+1)=$

$(x+1)(x^3 - 4x^2 + 4x -45)$.

And the find $x^3 - 4x^2 + 4x -45=0$ for integer $x$ would mean

$x^3 + 4x = 4x^2+45$ would require $x>0$ and $x|45$ simple trial would have $x=5$. So $x-5$ is a factor and

$x^3 - 4x^2 + 4x -45 = x^2(x-5) +x^2 +4x -45=$

$x^2(x-5) + x(x-5) + 9x - 45=$

$x^2(x-5) + x(x-5) + 9(x-5)=$

$(x-5)(x^2 +x + 9)$

So $x^4-3x^3 -41x -45=(x+1)(x-5)(x^2 + x +9)$.

No by the rational root theorem the rational roots of $x^2 + x +9$ must be $x =\pm 1, \pm 3$ and it's obvious none of those work. (Also by the quadratic formula $x =\frac {-1\pm\sqrt{1^2 -4*9}}2$ is not real, let only rational.)

So that's the complete factorization $x^4-3x^3 -41x -45=(x+1)(x-5)(x^2 + x +9)$