To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into $\displaystyle ax^2 + bx + c$ quadratic form.

$\displaystyle 8x - 16 = x^2$ ---> $\displaystyle x^2 - 8x + 16 = 0$

First, create two blank binomials.

$\displaystyle x^2 -8x +16 = (\_ + \_)(\_ + \_)$

Start by factoring our first term back into the first term of each biniomial. Since the only reasonable roots of $\displaystyle x^2$ are $\displaystyle x$ and $\displaystyle x$, we know that

$\displaystyle x^2 -8x +16 = (x + \_)(x+ \_)$

Next, factor out our constant, ignoring the sign for now. The factors of $\displaystyle 16$ are either $\displaystyle 1$ and $\displaystyle 16$, $\displaystyle 2$ and $\displaystyle 8$, or $\displaystyle 4$ and $\displaystyle 4$, We must select those factors which have either a difference or a sum equal to the value of $\displaystyle b$ in our trinomial. In this case, neither $\displaystyle 2$ and $\displaystyle 8$ nor $\displaystyle 1$ and $\displaystyle 16$ can sum to $\displaystyle 8$, but $\displaystyle 4$ and $\displaystyle 4$ can. Now, we can add in our missing values:

$\displaystyle x^2 -8x +16 = (x + 4)(x+ 4)$

One last step remains. We must check our signs. Since $\displaystyle c$ is positive in our trinomial, Either both signs are negative or both are positive. To figure out which, check the sign of $\displaystyle b$ in our trinomial. Since $\displaystyle b$ is negative, both signs in our binomials must be negative.

$\displaystyle x^2 -8x +16 = (x - 4)(x - 4)$

Thus, our two binomial factors are $\displaystyle (x -4)$ and $\displaystyle (x-4)$. Note that this can also be written as $\displaystyle (x-4)^2$, and if you graph this, you get a result identical to $\displaystyle x^2 - 8x + 16$.

To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into $\displaystyle ax^2 + bx + c$ quadratic form, but this is already done. Note that this is a special trinomial: $\displaystyle b=0$, and thus only $\displaystyle ax^2$ and $\displaystyle c$ are present to be manipulated.

$\displaystyle x^2 - 25$

First, create two blank binomials.

$\displaystyle x^2 - 25 = (\_ + \_)(\_ + \_)$

Start by factoring our first term back into the first term of each biniomial. Since the only reasonable roots of $\displaystyle x^2$ are $\displaystyle x$ and $\displaystyle x$, we know that

$\displaystyle x^2 -25 = (x + \_)(x+ \_)$

Next, factor out our constant, ignoring the sign for now. The only factors of $\displaystyle 25$ are either $\displaystyle 1$ and $\displaystyle 25$, or $\displaystyle 5$ and $\displaystyle 5$, We must select those factors which have either a difference or a sum equal to the value of $\displaystyle b$ in our trinomial. In this case, only $\displaystyle 5$ and $\displaystyle 5$ sum to $\displaystyle 0$, which we use since $\displaystyle b$ is not present.

$\displaystyle x^2 -25 = (x + 5)(x + 5)$

One last step remains. We must check our signs. Since $\displaystyle c$ is negative in our trinomial, one sign is positive and one is negative.

$\displaystyle x^2 -25 = (x + 5)(x - 5)$

Thus, our two binomial factors are $\displaystyle (x +5)$ and $\displaystyle (x-5)$.

This is called a difference of squares. If you see a trinomial in the form $\displaystyle x^2 - c^2$, the roots are always $\displaystyle (x + c)$ and $\displaystyle (x-c)$.

To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into $\displaystyle ax^2 + bx + c$ quadratic form. It isn't ideal to work with $\displaystyle a \neq 1$, but $\displaystyle a$ in this equation cannot be reduced.

$\displaystyle -2x^2 = x-3$ ---> $\displaystyle 2x^2 + x - 3 = 0$

First, create two blank binomials.

$\displaystyle 2x^2 + x - 3 = (\_ + \_)(\_ + \_)$

Start by factoring our first term back into the first term of each biniomial. Since the only reasonable roots of $\displaystyle 2x^2$ are $\displaystyle 2x$ and $\displaystyle x$, we know that

$\displaystyle 2x^2 + x - 3 = (2x + \_)(x+ \_)$

Next, factor out our constant, ignoring the sign for now. The factors of $\displaystyle 3$ are $\displaystyle 3$ and $\displaystyle 1$. Note that these do not normally sum or difference to $\displaystyle 1$, but the presence of $\displaystyle 2x$ as a term means one term will be doubled in value when creating $\displaystyle b$. This means we either get $\displaystyle 6$ and $\displaystyle 1$ (no good), or $\displaystyle 2$ and $\displaystyle 3$ (good). Remember to add the number we want to double inside the binomial opposite $\displaystyle 2x$, so it multiplies correctly. Now, we can add in our missing values:

$\displaystyle 2x^2 + x - 3 = (2x + 3)(x + 1)$

One last step remains. We must check our signs. Since $\displaystyle c$ is negative in our trinomial, one sign is positive and one is negative. To figure out which, check the sign of $\displaystyle b$ in our trinomial. Since $\displaystyle b$ is positive, the bigger of our two terms after any multipliers are completed is the positive value. Since our terms after mutiplying are $\displaystyle 2$ and $\displaystyle 3$, the $\displaystyle 3$ must be the positive term.

$\displaystyle 2x^2 + x - 3 = (2x + 3)(x - 1)$

Thus, our two binomial factors are $\displaystyle (2x+3)$ and $\displaystyle (x-1)$.

To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into $\displaystyle ax^2 + bx + c$ quadratic form, but this is already done. It isn't ideal to have $\displaystyle a \neq 0$, but we cannot reduce the constants further.

$\displaystyle 6x^2+ 5x -6$

First, create two blank binomials.

$\displaystyle 6x^2+ 5x -6 = (\_ + \_)(\_ + \_)$

Start by factoring our first term back into the first term of each biniomial. There are two valid factorings of $\displaystyle 6x^2$. We could have $\displaystyle 6x \cdot 1x$ or $\displaystyle 2x \cdot 3x$, and there's not really a valid system for figuring out which is correct (unless you use the quadratic formula). As a loose rule of thumb, if $\displaystyle b$ is near in value to $\displaystyle a$, it's more likely (though by no means necessary) that the factors of $\displaystyle a$ are also close in value. So, let's try $\displaystyle 2x \cdot 3x$.

$\displaystyle 6x^2+ 5x -6 = (3x + \_)(2x+ \_)$

Next, factor out our constant $\displaystyle c$, ignoring the sign for now. The factors of $\displaystyle 6$ are either $\displaystyle 1$ and $\displaystyle 6$, or $\displaystyle 2$ and $\displaystyle 3$, but the presence of $\displaystyle 2x$ and $\displaystyle 3x$ as terms means one term will be doubled in value and the other will be tripled when creating $\displaystyle b$. So, for $\displaystyle 1$ and $\displaystyle 6$ we either produce $\displaystyle 2$ and $\displaystyle 18$ (no good), or $\displaystyle 3$ and $\displaystyle 12$ (no good) for $\displaystyle b$. If we try $\displaystyle 2$ and $\displaystyle 3$ with $\displaystyle 2x$ and $\displaystyle 3x$, we get either $\displaystyle 6$ and $\displaystyle 6$ (no good) or $\displaystyle 4$ and $\displaystyle 9$ (good!) Remember to add the numbers we want to use opposite the $\displaystyle ax$ term we want to combine them with , so it multiplies correctly. 

Now, we can add in our missing values:$\displaystyle 6x^2+ 5x -6 = (3x + 2)(2x+ 3)$

One last step remains. We must check our signs. Since $\displaystyle c$ is negative in our trinomial, one sign is positive and one is negative. To figure out which, check the sign of $\displaystyle b$ in our trinomial. Since $\displaystyle b$ is positive, the bigger of our two terms after any multipliers are completed is the positive value. Since our terms after mutiplying are $\displaystyle 9$ and $\displaystyle 4$, the number that becomes $\displaystyle 9$ must be the positive term.$\displaystyle 6x^2+ 5x -6 = (3x - 2)(2x+ 3)$

Thus, our two binomial factors are $\displaystyle (3x -2)$ and $\displaystyle (2x +3)$.

Through factoring, the sum of the two roots must equal 2, and the product of the two roots must equal $\displaystyle \textup{-3}$. $\displaystyle \text{-1}$, and 3 satisfy both of these criteria.

Given the following expression:

$\displaystyle x^{2}+12x + 32$

We need to find factors of $\displaystyle 32$ that add up to $\displaystyle 12$. 

$\displaystyle 32$ can be broken down into the following factors:

$\displaystyle 32 , 1$

$\displaystyle 16, 2$

$\displaystyle 8,4$

Of these choices, only $\displaystyle 8 + 4$ adds up to $\displaystyle 12$. Additionally, the coefficient in front of the variable is $\displaystyle 1$, so we do not need to worry about that when finding these values. There are no negatives in the quadratic expression, so the signs in the factored form are all positive. This gives us the final answer of

$\displaystyle (x+8)(x+4)$

You can use the FOIL method to re-expand the expression and check your work!

To factor a quadratic equation in the form

$\displaystyle ax^{2}+bx+c$, where $\displaystyle a=1$, find two integers that have a sum of $\displaystyle b$ and a product of $\displaystyle c$.

For this equation, that would be 9 and 5.

$\displaystyle 9+5=14$

$\displaystyle 9\cdot 5=45$

Therefore, the solutions to this equation are $\displaystyle -9$ and $\displaystyle -5$.

$\displaystyle x^{2}+3x=88$

Begin by setting the equation equal to 0 by subtracting 88 from both sides.

$\displaystyle x^{2}+3x-88=0$

Now that the equation is in the form $\displaystyle ax^{2}+bx+c=0$, find two integers that sum to $\displaystyle b$ and have a product of $\displaystyle c$.

For this equation, those integers are $\displaystyle 11$ and $\displaystyle -8$.

$\displaystyle 11-8=3$

$\displaystyle 11\cdot -8=-88$

Therefore, the solutions to this equation are $\displaystyle (-11,8)$

$\displaystyle n^{2}+17n-30=-102$

Begin by setting the equation equal to zero by adding 105 to each side.

$\displaystyle n^{2}+17n+72=0$

For an equation in the form $\displaystyle ax^{2}+bx+c=0$, where $\displaystyle a=1$, find two integers that have a sum of $\displaystyle b$ and a product of $\displaystyle c$.

For this equation, that would be 8 and 9.

$\displaystyle 8+9=17$

$\displaystyle 8\cdot 9=72$

Therefore, the solutions to this equation are $\displaystyle (-8,-9)$

Rewriting the equation as $\displaystyle \small 76a^3+19a^2+16a+4=0$, we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is $\displaystyle \small 19a^2$ and between the third and fourth terms, the GCF is 4. Thus, we obtain $\displaystyle \small \small 19a^2(4a+1)+4(4a+1)=0 \rightarrow (4a+1)(19a^2+4)=0$.   Setting each factor equal to zero, and solving for $\displaystyle \small a$, we obtain $\displaystyle \small a=-\frac{1}{4}$ from the first factor and $\displaystyle \small a^2=-\frac{4}{19}$ from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept $\displaystyle \small a=-\frac{1}{4}$.