Why is every factor of a polynomial equal to 0?

Question

When factoring any polynomial, it's generally agreed upon (at least in my math class) that all the factors you get are equal to $0$ (When you solve). I haven't really been able to myself build a serious argument against until recently, when we were introduced to synthetic division, and therefor, given factors.

Take for example, this problem which we did in class today.

Factor: $f(x)=2x^3+3x^2-39x-20$ Given $f(4)=0$

Because $f(4)=0$, we know that one factor is $(x-4)$, and using synthetic division to solve the rest, we get factors of $(x-4)(2x^2+11x+5)$, and because we're solving, $(x-4)(2x^2+11x+5)=0$.

This is the point where I get confused. Because we know $x-4 = 0$, a multiplication by 0 always equals 0, how can we know that $2x^2+11x+5=0$? Couldn't $2x^2+11x+5 = \frac{35}{\pi(x-56)}$, because $\frac{35}{\pi(x-56)} \times 0 = 0$?

Why can we assume $(2x^2+11x+5)=0$, instead of anything else?

Answer 1

Making an answer out of Compulsive's comment...

Using the hint $f(4) = 0$, you found $$ f(x) = (x-4)(2x^2 + 11x +5). $$ Now, you are trying to find all numbers $x$ for which $f(x) = 0$. Assume for a second that you have found a number $x$ such that $f(x) = 0$. Because $f(x)$ is the product of $x-4$ and $2x^2 + 11x +5$, you have either $$x-4 = 0$$ or $$2x^2 + 11x +5 = 0$$ From there, considering the two possibilities, you can find what $x$ must be to indeed satisfy the equation $f(x) = 0$.

Answer 2

We assume/hope that we will be able to factor $2x^3 + 3x^2-39x-20$ as $2(x+a)(x+b)(x+c)$. If $f(4) =0$ then we know that $2(4+a)(4+b)(4+c)=0$. But that can only happen if one of those terms $4+a, 4+b, $ or $4+c$ is equal to zero. So we know that either $a, b,$ or $ c$ that one of them must be $4$. So let's say $c = -4$.

So we hope that we will be able to factor $2x^3 + 3x^2 - 39x - 20 = 2(x-4)(x+a)(x+b)$. By synthetic division we get that $2(x+a)(x+b) = (2x^2+11x+5)$ and we want to factor that.

Well we can use the quadratic formals the solve that if $2x^2 + 11x + 5 = 0$ then $x = \frac{-11 \pm \sqrt{121-4*5*2}}{4} = \frac {-11 \pm 9}{4} = -5; -1/2$.

So we know $f(-5) = 0$ so we know $2(-5 -4)(-5+a)(-5+b) = 0$. So that means either $-5 + a = 0$ or $-5 +b = 0$. That means either $a = 5$ of $b = 5$. So let's say $a = 5$ and we know that:

$f(x) = 2(x-4)(x+5)(x +a)$.

And we know $f(-1/2) = 0$ so we know $2(-1/2 -4)(-1/2+5)(-1/2 + a) = 0$ so $a = 1/2.

And we know that $f(x) = 2(x-4)(x+5)(x+ 1/2)$.

Answer 3

Suppose you polynomial $p(x)$ can that can be factored into be factored into $(x-a)q(x)$

$p(a) = 0$ because $(a-a)q(a) = 0\cdot q(a) = 0$

There is no information here what $q(a)$ equals. It could be anything. It certainly doesn't have to be $0.$

and if $p(b) = 0$ and $b\ne a$ then $(b-a)q(b) = 0$

$(b-a)$ definitely does not equal $0$ and $q(b)$ must equal $0.$

Synthetic division.

Suppose you have some polynomial $p(x)$ (different from the one above) and you want to find $p(a).$

you attempt to divide $p(x)/(x-a)$ but it has a remainder $r$

Then $p(x) = (x-a)q(x) + r$

$p(a) = (a-a)q(a) + r = r$

Answer 4

It is not quite clear what you are asking. But this is how factors work -- hopefully this will clear up whatever your problem is. I hope I'm addressing what you're after. It really is hard to get what you are asking.

When you have a polynomial like, say $2x^3 + 3x^2 - 39x - 20$, none of its factors are 0. If any where, then the polynomial would be identically 0, so there is actually only one polynomial with $0$ factors, that is, the zero polynomial, $\mathbf{0}$ (bolded to make distinct from numbers).

What happens when you have a number like $x = 4$ that makes the polynomial zero when evaluated at that particular $x$-value, that means that there is a factor who is 0 at that particular number for $x$, because if you write the polynomial in factored form, you are writing it as a product, and for a product to equal zero at least one term must also be zero, thus at least one factor must equal zero at that particular $x$-value. The lowest-possible-degree such factor would be (in this case) $(x - 4)$. The trick is you should not confuse being 0 at a particular $x$-value with being identically zero, that is, equal to the zero polynomial as a polynomial, which no factor of any (nonzero) polynomial is. So we pull that out by long division. The other factor is not going to be zero in zero-polynomial sense, and it may not be 0 even at our specific $x$-value where the other is. If it is, we can then factor out another term of the form $(x - a)$, and this means $a$ is a root of higher order. In general, if we can factor out $(x - a)$ $n$ times and no more, then the root $a$ has order $n$.

So in your example, you can pull out $(x - 4)$ once, and this factor is 0 at $x = 4$, but your other factor $(2x^2 + 11x + 5)$ evaluates to 81 there (which tells us that $x = 4$ is a root of order 1 since we cannot pull out another factor of $(x - 4)$ from the leftover part.). So not all factors equal zero here.

But if we took the polynomial

$$x^3 - 11x^2 + 40x - 48$$

we would factorize it into

$$(x - 4)(x^2 - 7x + 12)$$

but now both factors do go to zero at $x = 4$, thus $x = 4$ must be of order greater than 1. We can then factorize another $(x - 4)$ out of the quadratic to get the final factorization as

$$(x - 4)(x - 4)(x - 3)$$

or

$$(x - 4)^2 (x - 3)$$

showing that $x = 4$ has order 2, and the other root is $x = 3$, of order 1. Note also that now, in the fully factorized form, only the first factor (or two factors) goes to 0 at $x = 4$, the other does not.

So it is in general not true that "every factor of a polynomial is zero when solving". They are never identically zero except in one case alone, and in general, even where one factor goes to zero at a specific value of $x$ that is a root, the others need not necessarily go to zero themselves at that same place.