$ax^2+bx+c$ onde $ac<0$

Factoring Quadratics - $ax^2+bx+c, ac<0$

Here is another way to evaluate a common quadratic form with factor by grouping.

The first step to factor a polynomial with the form $ax^2+bx+c$ when $ac<0$ is to find two numbers, $d$ and $e$, where $d\cdot e=ac$ and $d+e=b$. $d$ and $e$ are the middle terms. Assume that we have found the two numbers. Since $ac$ is negative, we need to set $-d \cdot e=ac$ and $-d+e=b$.

Let's plug in $d$ and $e$. Remember that we picked $-d$, so the polynomial becomes $ax^2-dx+ex+c$. This can be grouped with parentheses to $(ax^2-dx)+(ex+c)$. These groups share a greatest common factor. The GCF gets factored out, and we are left with an expression of two multiplied binomials. This is the factored form of the quadratic.

Example 1:

$$-2x^2 + 5x + 3$$

Since $-1 ⋅ 6 = -6 = -2 ⋅ 3$, and since $-1 + 6 = 5$, we can rewrite our polynomial as follows:

$$-2x^2 - 1x + 6x + 3$$

Now, we factor out greatest common factors from our polynomial's first two terms and from our polynomial's last two terms:

$$-x(2x + 1) + 3(2x + 1)$$

Finally, we factor the binomial $2x + 1$ out of our expression (note that we are only able to do this because our expression consists of two terms that are both being multiplied by the binomial $2x + 1$):

$$(-x + 3)(2x + 1)$$ $$= -(x - 3)(2x + 1)$$

Here's a graphic with another example!

Test Yourself!

Let's do some more problems.

Let's start with a factorization where $a=1$ to get into the factoring mindset.

When factoring a trinomial ($3$ terms) in the standard form $ax^2+bx+c$ it can be more challenging than factoring a trinomial in the form $x^2+bx+c$.

If you multiplied $(-4x+1)(2x+8)$

Using a FOIL or table method, you would get $-8x^2-30x+8$.

Since factoring is the reverse of multiplication, you have to think how to multiply to get $-8x^2$. You would multiply either $-4x$ and $2x$, $4x$ and $-2x$, $8x$ and $-x$, or $-8x$ and $x$. Due to this, it is trickier to factor polynomials when $ac<0$ and $a$ is not $1$ because using a trial and error method, there is more than one option for the coefficient of $x$.

Follow these steps to factor a trinomial $-ax^2+bx+c$

1. Check to see if there is a GCF between the terms.

2. Use a table with $4$ sections ($2$ by $2$ dimensions). Fill in the $x^2$ term in the top left and the constant term in the bottom right.

3. Multiply $a$ and $c$.

4. Find factors of $ac$ that add up to $b$.

5. Fill in the empty boxes putting an $x$ after the numbers.

6. Look at each row and factor out the common factor.

7. Divide the terms by the GCF to get the other binomial factor.

1) Check to see if there is a GCF between the terms

For example, to factor the polynomial $-2x^2-16x-24$ you would first notice that you could factor out or divide each term by $-2$ giving you

$-2\left(x^2+8x+12\right)$

Then, it is easier to factor and you would notice that $6$ and $2$ multiply to $12$ and add up to $8$. Don't forget to leave the $-2$ as a factor.

The fully factored form would be: $-2\left(x+6\right)\left(x+2\right)$

If there is no common monomial factor, as there often is not, that is ok! You just always want to check.

2) Use a table with $4$ sections ($2$ by $2$ dimensions). Fill in the $x^2$ term in the top left and the constant term in the bottom right.

To factor $-6x^2+4x+2$ fill in the boxes as shown below:

3) Multiply a and c.

In this example, $a$ is $-6$ and $c$ is $2$. So $ac = -12$.

4) Find factors of ac that add up to b.

Since $ac = -12$ and $b = 4$

$-2$ and $6$ multiply to $-12$ and add up to $4$.

5) Fill in the empty boxes putting an $x$ after the numbers, as shown.

6) Look at each row and factor out the common factor.

For $-6x^2+6x$, the GCF is $\color{green}{6x}$.

For $-2x+2$, the GCF is $\color{purple}{2}$.

Therefore: $(\color{green}{6x}+\color{purple}{2})$ is one of the factors.

7) Divide the terms by the GCF to get the other binomial factor.

$6x^2$ divided by $-6x$ is $-x$.

$6x$ divided by $6x$ is $1$.

$\left(-x+1\right)$ is the other factor.

When fully factored, $-6x^2+4x+2 = \left(-x+1\right)\left(6x+2\right)$.

(Video) Example: Factoring a Quadratic with a Negative Leading Coefficient

by Global Freshman Academy

This video by Global Freshman Academy discusses how to factor a quadratic with a negative leading coefficient.

$\large{\text{Summary}}$

Let's start with a problem. You start to factor by pulling a negative out front. $$-2x^2-3x+14=-(2x^2+3x-14)$$ Now, we find the factors that multiply to get $-28$ and add to $3$. We can use $-4$ and $3$. We see, $$-(2x^2+3x-14)=-(-2x^2-4x+7x-14)$$ We now look for common factors in $2x^2-4x$ and $7x-14$. We see we can pull out a $2x$ and a $7$. Therefore, our factors are $-(x-2)$ and $(2x+7)$.

You can check your work by using the FOIL method. Our final answer is $$-2x^2-3x+14=-(x-2)(2x+7)$$

Example $2$: Start with the quadratic and pull out a common factor. $$-4x^2+28x-48 = -4(x^2-7x+12)$$ Try this problem from here and see if you get the correct answer!

The answer is $$-4x^2+28x-48 = -4(x-3)(x-4)$$