Question 277475
{{{p(x) = 3(2x+7)^2(x-1)^2-(2x+7)(x-1)^3}}}
To find zeros we need to factor the polynomial. The straightforward way to factor this would be to simplify it first. But this is a lot of work. We would have to square 2x+7 and both square and cube x-1, add the like terms and then try to factor what we end up with.<br>
But you may notice that there are already common factors on each side of the subtraction in the middle. We can use this to our advantage and make the problem much easier. First I am going to rewrite the expression to make what I do clearer:
{{{p(x) = 3(2x+7)red((2x+7)(x-1)^2)-red((2x+7)(x-1)^2)(x-1)}}}
The GCF of the expressions on each side of the subtraction are in red. I will now factor out this GCF from each side:
{{{p(x) = red((2x+7)(x-1)^2)(3(2x+7)-(x-1))}}}
Now I'll simply expression in the last factor:
{{{p(x) = (2x+7)(x-1)^2(6x+21-(x-1))}}}
{{{p(x) = (2x+7)(x-1)^2(6x+21-x+1)}}}
{{{p(x) = (2x+7)(x-1)^2(5x+22)}}}
And we now have p(x) factored. And the zeros will be the values of x that make the factors zero:
2x+7 = 0  or  x-1 = 0 or 5x+22 = 0
Solving these we get:
x = -7/2 or x = 1 or x = -22/5
These are the zeros of p(x).