Attempting to factor $(x^2)+5x=0$ via the grouping method.

Question

Now, I am perfectly aware that I can easily factor this by taking out the common factor, so I would get $x(x+5)$. This way the roots of my equation would be $0$ and $-5$ What is confusing me is that, even though it is a longer and a redundant way, the grouping method should work. I am referring to the quadratic equation in the form of $(ax^2)+bx+c$ there should be factors ac which has a sum equal to $b$. Of course, I can also re-write the original equation as $(x^2)+5x+0$ since my c would equal to $0$ in this case. Now, I need to find two numbers that add up to 5 that have a product of $0.$ It is obvious that these numbers are $0$ and $5.$ But doesn't this indicate that the factored form should be $x(x-5)=0$?

I think I made a mistake somehwere, I just can't figure out where...

Answer 1

The roots $r_1$ and $r_2$ of $x^2+\alpha x+\beta$ are such that $r_1r_2=\beta$ and that $r_1+r_2=\color{red}{\mathbf-}\alpha$. So, you're after two numbers with product $0$ and sum $-5$; these are $0$ and $-5$.

Answer 2

Only your last statement is wrong. You're right that if you want to use the grouping method (also called splitting the middle term, the AC method, and possibly other names) the numbers should be $5$ and $0$. But remember, from there you split the linear term and get $$x^2+5x+0x+0=x(x+5)+0(x+5)=(x+0)(x+5)=x(x+5)$$ However, as you said your first method is easier in this case.

Answer 3

$ax^2 + bx + c = x^2 + 5x + 0 = 0.$

This will have roots

$\displaystyle \frac{1}{2a}\left[-b \pm\sqrt{b^2 - 4ac}\right].$

This equals

$\displaystyle \frac{1}{2}\left[-5 \pm\sqrt{25}\right].$

This simplifies to the two roots of

$\left[\frac{1}{2} \times 0\right]$ and $\left[\frac{1}{2} \times -10\right].$