Why are these two given expressions equivalent? $6\cos^2(\frac {x}{2})-7\cos(\frac {x}{2})+2=0$ is equivalent to the expression $(3\cos(\frac {x}{2})-2)(2\cos(\frac {x}{2})-1) = 0$
Could someone explain the steps involved in evaluating the first expression to equal the second expression?
I noticed the expression in this question here, and I was curious how this could be achieved.
 A: Key Idea: Knowing the roots of a polynomial let us factor it. So if $r_1, r_2,\ldots,r_n$ are all the roots of a given polynomial, then we can factorize it as: $a(x-r_1)(x-r_2)\cdots(x-r_n)$, where $a$ is the coefficient of the term with $x^n$.
Notice: $$6\,\color{red}{\cos\left(\tfrac x2\right)}^2+7\,\color{red}{\cos\left(\tfrac x2\right)}+2=0.$$ So we can get a simple quadratic equation by letting $t=\cos\left(\tfrac x2\right)$, then our equation becomes: $$6t^2-7t+2=0.\tag{$\star$}$$
Using the quadratic formula we get: $$6t^2-7t+2=0\iff t=\dfrac23\quad\color{grey}{\text{or}}\quad t=\dfrac12.$$
So we can write $(\star)$ as: $$6\left(t-\dfrac23\right)\left(t-\dfrac12\right)=0\iff (3t-2)(2t-1)=0.$$
Now replace $t$ with our original substitution.
Side note: This $t$-substitution isn't necessary at all, when you noticed that you had an expression of the form $a(\text{something})^2+b(\text{something})+c=0$ you can directly use the quadratic formula to solve for that $\text{something}$.
A: We need find $pq,$ such that $pq=6\cdot2=12;  p+q=7$
$$6a^2-7a+2=6a^2-(4+3)a+2=2a(3a-2)-(3a-2)=(3a-2)(2a-1)$$
A: The answer comes from factoring.
When you foil the two parentheticals from the second expression, their products can be seen to equal the first.
We treat $\cos(\frac {x}{2})$ much like we would the variable $x$ within any given expression.
If given the expression $2x^2-3x+1 = 0$, we are able to factor this to show that $(-2x+1)(-x+1) = 0$.
Given this particular expression: $6\cos^2(\frac {x}{2})-7\cos(\frac {x}{2})+2=0$, we are able factor this to show that $(3\cos(\frac {x}{2})-2)(2cos(\frac {x}{2})-1) = 0$.

When foiled, it is shown that:
First: $3\cos(\frac {x}{2}) \cdot (2\cos(\frac {x}{2}) = 6\cos^2(\frac {x}{2})$
Second: The product of $3\cos(\frac {x}{2}) \cdot -1 = -3\cos(\frac {x}{2})$ and the product of $2cos(\frac {x}{2}) \cdot -2 = -4cos(\frac {x}{2})$, and their sum equals $-7\cos(\frac {x}{2})$
Third: $-2 \cdot -1 = 2$
Final, we add the first, second, and third steps together, and their sums are show to be equivalent to the initial expression: $6\cos^2(\frac {x}{2})$ [first] $-7\cos(\frac {x}{2})$[second]$+2$[third]$=0$.

And the result is [Final].
