I was wondering if someone could clear this question I had.

$$x^2+4x-21=0$$ Factoring it out: $$(x+7)(x-3)$$ How does that mean that $x=-7$ or $x=3$?

How is it that we can just say "Alright, let's just forget one part of the equation $(x-3)$, and solve for $x$, or forget $x+7$ and solve for $x$," and it's going to hold true?

Thanks!!

• Should it be $(x + 7)(x - 3) =0$? – Azlif Jan 10 '18 at 14:55
• Your question probably is $x^2 - 4x - 21 = 0$ . If not, then you factored it wrongly. – Manish Kundu Jan 10 '18 at 14:58
• This is because $0\cdot \text{anything}=0$ so if $x-3=0$ we can forget the rest part $(x+7)$ because no matter what the value of the rest part is the value of the expression will be $0$. – kingW3 Jan 10 '18 at 15:23
• Thanks to everyone, for all the comments and answers :) – user472288 Jan 10 '18 at 16:25

Firstly, you probably meant $x^2 - 4x - 21 = 0$ as the equation $x^2 + 4x - 21 = 0$ does not have the factors you mentioned.
Now you have: $(x+3)(x-7) = 0$ . When the product of two numbers is zero, then either or both of them is zero, which means that,
$(x+3) = 0$, which implies that $x = -3$ , or $(x-7) = 0$ , i.e, $x = 7$ .
• Manish, I was wondering if you could clear this up for me: It means that one part has to equal 0 like you described. But we have to plug in an $x$ that makes this true right? If we do, then say $x=3$ so that $(x-3)=0$. But then we have to plug into $3$ into the other side as well, to make it $(x+7)=10? – user472288 Jan 10 '18 at 15:12 • No, you don't need to plug in anything. You simply equate the factors with zero and get the corresponding values of$ x $– Manish Kundu Jan 10 '18 at 15:13 Prove that if$ab=0$, then$a=0$or$b=0$. solving $$x^2+4x-21=0$$ we get $$x_{1,2}=-2\pm\sqrt{25}$$ and form here we get..... we get$x_1=-7,x_2=3$so we have $$(x+7)(x-3)=x^2+4x-21$$ The technical term is the Zero Factor Theorem. It states that if two (or more) numbers multiply to make zero, then at least one of them had to already be zero - in other words, if I multiply together a bunch of things that aren't zero, I can't get zero out. In this case, factoring tells us that$x^2 + 4x - 21$is the same as$(x - 7)(x + 3)$. So if$x^2 + 4x - 21 = 0$, then we know that$(x - 7)(x + 3) = 0$. By the Zero Factor Theorem, that means we have one of two situations: either$x - 7 = 0$or$x + 3 = 0$. If$x - 7 = 0$, then$x = 7$. If$x + 3 = 0$, then$x = -3$. • @DietrichBurde Given the level of the question and the tags on it, I of course answered under the assumption that we're looking exclusively at real or complex numbers. – Reese Jan 10 '18 at 16:20 We have to be careful, in which domain we solve such an equation. For example, in$R=\mathbb{Z}/4$we have$2\cdot 2=0$, but none of the factors is zero. In a field$K$however we have that$ab=0$always implies that$a=0$or$b=0$. Hence $$(x+7)(x-3)=0$$ implies then that$x+7=0$or$x-3=0$. • Aren't these terms called complex numbers? – user472288 Jan 10 '18 at 15:03 • No, the terms are called "linear factors" of$x^2+4x-21$. – Dietrich Burde Jan 10 '18 at 16:31$x^2 +4x -21=0.x^2 +4x -21 = (x+7)(x-3) =0.$The product$(x+7)(x-3) = 0\iff $one of the factors$(x+7)$or$(x-3)$is 0. 1)$x+7= 0$gives$x=-7;$2)$x-3 = 0$gives$x=3.$Note: For real$a,b:ab = 0$implies$a=0$or$b=0$. ('Or' means : One factor or the other or both) Helps? This is how I learned it. Factoring an equation gives you the answers, as to where on the curve is$f(x)=0$. So when you factorize an equation,$x^2+4x−21=0$, you want to find which value of$x$satisfies$f(x)=0$. In this case your equation was$x^2+4x−21=(x+7)(x-3)$. Then you consider each bracket separately,$x+7=0$and$x-3=0$,$x_1=-7, x_2=3$. So your answers are$x_1=-7, x_2=3$. You can consider each bracket individually because if one of the brackets ends up as a$0$then$f(x)$will be$0\$.