Solving polynomial equations

Example 1: Solve $x^2-16x+28=0$

Solution: Multiplying the coefficient of $x^2$ and the constant term, $1\cdot 28 = 28$. Now we need to find two numbers whose product is $28$ and whose sum is $-16$. The two numbers are $-2$ and $-14$. Now replace $-16x$ with $-2x-14x$:

$x^2-2x-14x+28=0$

Now, factor by grouping, factor the first two terms and next the last two terms:

$x(x-2)-14(x-2)=0$

Taking the common factors out,

$(x-2)(x-14)=0$

Applying the zero product rule,

$x-2=0$ or $x-14=0$

Solving for $x$,

$x=2$ or $x=14$

Example 2: Solve $8w^2-32w=0$

Solution: In this equation there is no constant term, so just factor out the greatest common factor and apply the zero product rule:

$\begin{align*} 8w^2-32w &=0\\ 8w(w-4) &=0\\ \end{align*}$

Applying the zero product rule,

$w=0$ or $w-4=0$.

or

$\boxed{w=0 \: or\: w=4}$

Example 3: Solve $4u^2+12u=-9$

Solution: This equation is not in the form $ax^2+bx+c=0$, so first write it in that form. Move the number on the right-hand side to the left by adding $9$:

$4u^2+12u+9=0$

The first and last terms can be written as perfect squares:

$(2u)^2+12u+3^2=0$

This looks like the form $a^2+2ab+b^2$. To check, compare it with our equation: we have $a=2u$ and $b=3$. Taking the product $2ab$, we get $2\cdot2u\cdot 3=12u$, which is the same as the middle term. Therefore, the quadratic is exactly of the form $a^2+2ab+b^2$, so we can write it as $(a+b)^2$. Therefore,

$(2u+3)^2=0$

or

$(2u+3)(2u+3)=0$

Applying the zero product rule,

$2u+3=0$ or $2u+3=0$

Since the two equations above are the same, take one of them and solve for $u$, and you will get

$\boxed{u=-3/2}$