First, make one side zero. Subtracting 6 we get:
A number that makes a whole polynomial equal to zero is called a root. So we are interested in the roots of this polynomial.
If some number, let's call it "r", is a root of a polynomial then (x-r) will be a factor and vice versa. So if we can factor the polynomial, each factor of the form (x-r) will tell us a root.
So we set out to factor the polynomial. The greatest common factor is 1 (which we rarely bother to factor out). There are too many terms for the factoring patterns or for trinomial factoring. And since I do not see how to factor this by grouping, we are left with factoring by trial and error of the possible rational roots.
The possible rational roots of a polynomial are all the ratios, positive and negative, which can be formed using a factor of the constant term on top and a factor of the leading coefficient on the bottom. The constant term (at the end) is 6. (Actually the term is -6 but since we're going to include all positive and negative ratios anyway, it doesn't matter if we just use 6.) The factors of 6 are 1, 2, 3 and 6. The leading coefficient (at the front) is a 1 (whose factors are 1's). So the possible rational roots are:
+1/1, +2/1, +3/1 and +6/1
which simplify to:
+1, +2, +3 and +6
Now we try these possible roots to see if any of them are actually roots. 1 and -1 can be checked with mental math since powers of these numbers are pretty easy.
We should find that 1 is actually a root. This also means that (x-1) is a factor of the polynomial. And if it's a factor of the polynomial then it will divide evenly into it. We'll use synthetic division:
1 | 1 4 1 -6
---- 1 5 6
----------------
1 5 6 0
The remainder, in the lower right corner, is zero (as expected). The reason we divided is to find out what the other factor is. The rest of the bottom row tells us what the other factor is. The "1 5 6" translated into . So at this point, our equation with partially factored polynomial looks like:
The second factor is a trinomial that factors easily. (Note: We could continue to do trial and error of the other possible rational roots instead of factoring the trinomial. But it's more time-comsuming so I'll take the easy way out for now. At the end of the problem I will finish the factoring using the possible roots and synthetic division.)
We already know that one root is 1. The other two factors will tell us the other two roots. We can either:- Rewrite the other factors in (x-r) form:
(x+2) = (x-(-2))
(x+3) = (x-(-3))
and then just "read" the roots: -2 and -3. Or... - Use the Zero Product Property and set the factors equal to zero:
x+2 = 0 or x+3 = 0
and then solve.
Either way, we get roots of 1, -2 and -3. These are the values of x that make equal zero and therefore are solutions to
P.S. Here's how you could finish the factoring using the possible rational roots and synthetic division. We have found that the partially factored polynomial is So we are now just trying to factor . This has the same constant term and the same leading coefficient so it has the same possible rational roots:
+1, +2, +3 and +6
If you know Descartes' rule of signs, you would know that there can be no positive roots of . So there is no point in trying any of the positive roots. So try the possible root -1:
-1 | 1 5 6
--- -1 -4
-----------
1 4 2
The remainder is not zero. So x-(-1) does not divide evenly. It is not a factor and -1 is not a root. Let's try -2:
-2 | 1 5 6
--- -2 -6
-----------
1 3 0
The remainder is zero. So x-(-2) (or x+2) does divide evenly. (x+2) is a factor and -2 is a root. The other factor of is "1 3" (from the bottom row) which translates into (x+3).
Our factored polynomial now looks like:
(x-1)(x+2)(x+3)
from which we should be able to see that -3 is the remaining root.