how do you factor $x^2 +kx+40$ over the integer

Question

please please help me, I'm having a lot of troubles. I tried to use a^2+2ab+b^2 formula (like i was told) but that's where get lost. I understand that Factoring uses the opposite operation, but 40 cannot be square rooted.

Answer 1

When solving problems like this, I often use factoring by decomposition. We have our expression: $$x^2+kx+40$$ Now, $1\times 40=40$. We now need to find two factors of $40$ that add up to $k$. In this case, there are a lot of possibilities. The first two factors that pop up in my mind are $4$ and $10$. This adds up to $14$, therefore one of the solutions is $k=14$. Try other factors, e.g. $5$ and $8$. This adds up to $13$, therefore another solution is $k=13$. Yet another solution is $k=-22$ (factors: $-20$ and $-2$).

Don't believe me? Try to factor these expressions: $$x^2+41x+40$$ $$x^2-41x+40$$ $$x^2+22x+40$$ $$x^2-22x+40$$ $$x^2+14x+40$$ $$x^2-14x+40$$ $$x^2+13x+40$$ $$x^2-13x+40$$ Prepare to be amazed :-)

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Answers to above expressions I told you to try to factor: $$x^2+41x+40=(x+1)(x+40)$$ $$x^2-41x+40=(x-40)(x-1)$$ $$x^2+22x+40=(x+20)(x+2)$$ $$x^2-22x+40=(x-20)(x-2)$$ $$x^2+14x+40=(x+4)(x+10)$$ $$x^2-14x+40=(x-4)(x-10)$$ $$x^2+13x+40=(x+5)(x+8)$$ $$x^2-13x+40=(x-5)(x-8)$$

Answer 2

Hint. If we have an integer factorisation $$x^2+kx+40=(x-a)(x-b)$$ then $$ab=40\quad\hbox{and}\quad -a-b=k\ .$$ Can you use the first equation to find some (or all) possible values of $a,b$? Then substitute into the second equation to find possible values of $k$.

Answer 3

If we look at the determinant $$D=k^2-160$$ we see that to get an integer solution, $D=q^2$ (some $q$). Moreover, both $k-q$ and $k+q$ must be divisible by $2$ (solutions: $x_{1,2}=\frac{-k\pm \sqrt{D}}{2}$).

Thus, $$k^2-q^2=160$$ $$(k-q)(k+q)=160$$ The only ways how $160$ may be factorized by even numbers are: $2*80,4*40, 8*20, 10*16$. Therefore, e.g., $$k-q=2$$ $$k+q=80$$ $\Rightarrow$ $k=41$, the others $k=13,14,22$ follows in the same way.

E.g. for $k=13$, $x_{1,2} = \frac{-13\pm 3}{2} = \{-8,-5\}$. To get positive integer solutions, we just take the negative $k=-13,-14,-22,-41$.

EDIT: Taking factorization $5*32=160$ there is yet another integer solution for $k=\pm18.5$, $x=\mp 16$.