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# Factorization of Quadratic Expressions with Negative Coefficients

## Factor quadratics with positive and negative coefficients

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Factorization of Quadratic Expressions with Negative Coefficients

### Factorization of Quadratic Expressions with Negative Coefficients

Factor a quadratic trinomial when a = 1, b is negative and c is positive

Now let’s see how this method works if the middle coefficient is negative.

Factor \begin{align*}x^2 - 6x + 8\end{align*}.

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

When negative coefficients are involved, we have to remember that negative factors may be involved also. The number 8 can be written as the product of the following numbers:

\begin{align*}8 = 1 \cdot 8 \quad \quad \text{and} \quad \quad 1 + 8 = 9\end{align*}

but also

\begin{align*}8 = (-1) \cdot (-8) \quad \quad \text{and} \quad \quad -1 + (-8) = -9\end{align*}

and

\begin{align*}8 = 2 \cdot 4 \quad \quad \text{and} \quad \quad 2 + 4 = 6\end{align*}

but also

\begin{align*}8 = (-2) \cdot (-4) \quad \quad \text{and} \quad \quad -2 + (-4) = -6.\end{align*}

The last option is the correct choice. The answer is \begin{align*}(x - 2)(x - 4)\end{align*}. We can check to see if this is correct by multiplying \begin{align*}(x - 2)(x - 4)\end{align*}:

\begin{align*}& \quad \quad \quad x - 2\\ & \underline{\;\;\;\;\;\;\;\;\;\;\;x - 4}\\ & \quad \ - \ 4x + 8\\ & \underline{x^2 - \ 2x\;\;\;\;\;\;\;}\\ & x^2 - \ 6x + 8\end{align*}

#### Factoring

1. Factor \begin{align*}x^2 - 17x + 16\end{align*}.

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

The number 16 can be written as the product of the following numbers:

\begin{align*}& 16 = 1 \cdot 16 && \text{and} && 1 + 16 = 17\\ & 16 = (-1) \cdot (-16) && \text{and} && -1 + (-16) = -17 \qquad (Correct \ choice)\\ & 16 = 2 \cdot 8 && \text{and} && 2 + 8 = 10\\ & 16 = (-2) \cdot (-8) && \text{and} && -2 + (-8) = -10\\ & 16 = 4 \cdot 4 && \text{and} && 4 + 4 = 8\\ & 16 = (-4) \cdot (-4) && \text{and} && -4 + (-4) = -8\end{align*}

The answer is \begin{align*}(x - 1)(x - 16)\end{align*}.

In general, whenever \begin{align*}b\end{align*} is negative and \begin{align*}a\end{align*} and \begin{align*}c\end{align*} are positive, the two binomial factors will have minus signs instead of plus signs.

Factor when a = 1 and c is Negative

Now let’s see how this method works if the constant term is negative.

2. Factor \begin{align*}x^2 + 2x - 15\end{align*}.

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;)(x\;\;\;\;\;)\end{align*}

Once again, we must take the negative sign into account. The number -15 can be written as the product of the following numbers:

\begin{align*}& -15 = -1 \cdot 15 && \text{and} && -1 + 15 = 14\\ & -15 = 1 \cdot (-15) && \text{and} && 1 + (-15) = -14\\ & -15 = -3 \cdot 5 && \text{and} && -3 + 5 = 2 \qquad \qquad (Correct \ choice)\\ & -15 = 3 \cdot (-5) && \text{and} && 3 + (-5) = -2\end{align*}

The answer is \begin{align*}(x - 3)(x +5)\end{align*}.

We can check to see if this is correct by multiplying:

\begin{align*}& \quad \quad \ \ x - \ 3\\ & \underline{\;\;\;\;\;\;\;\;\; x + \;5\;}\\ & \quad \quad 5x - 15\\ & \underline{x^2 - 3x\;\;\;\;\;\;\;\;}\\ & x^2 + 2x - 15\end{align*}

3. Factor \begin{align*}x^2 - 10x - 24\end{align*}.

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

The number -24 can be written as the product of the following numbers:

\begin{align*}& -24 = -1 \cdot 24 && \text{and} && -1 + 24 = 23\\ & -24 = 1 \cdot (-24) && \text{and} && 1 + (-24) = -23\\ & -24 = -2 \cdot 12 && \text{and} && -2 + 12 = 10\\ & -24 = 2 \cdot (-12) && \text{and} && 2 + (-12) = -10 \qquad (Correct \ choice)\\ & -24 = -3 \cdot 8 && \text{and} && -3 + 8 = 5\\ & -24 = 3 \cdot (-8) && \text{and} && 3 + (-8) = -5\\ & -24 = -4 \cdot 6 && \text{and} && -4 + 6 = 2\\ & -24 = 4 \cdot (-6) && \text{and} && 4 + (-6) = -2\end{align*}

The answer is \begin{align*}(x - 12) (x + 2)\end{align*}.

Factor when a = - 1

When \begin{align*}a = -1\end{align*}, the best strategy is to factor the common factor of -1 from all the terms in the quadratic polynomial and then apply the methods you learned so far in this section

4. Factor \begin{align*}-x^2 + x + 6\end{align*}.

First factor the common factor of -1 from each term in the trinomial. Factoring -1 just changes the signs of each term in the expression:

\begin{align*}-x^2 + x + 6 = -(x^2 - x - 6)\end{align*}

We’re looking for a product of two binomials in parentheses: \begin{align*}-(x\;\;\;\;)(x\;\;\;\;)\end{align*}

Now our job is to factor \begin{align*}x^2 - x - 6\end{align*}.

The number -6 can be written as the product of the following numbers:

\begin{align*}& -6 = -1 \cdot 6 && \text{and} && -1 + 6 = 5\\ & -6 = 1 \cdot (-6) && \text{and} && 1 + (-6) = -5\\ & -6 = -2 \cdot 3 && \text{and} && -2 + 3 = 1\\ & -6 = 2 \cdot (-3) && \text{and} && 2 + (-3) = -1 \qquad (Correct \ choice)\end{align*}

The answer is \begin{align*}-(x - 3)(x + 2)\end{align*}.

### Example

#### Example 1

Factor \begin{align*}x^2 + 34x - 35\end{align*}.

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

The number -35 can be written as the product of the following numbers:

\begin{align*}& -35 = -1 \cdot 35 && \text{and} && -1 + 35 = 34 \qquad (Correct \ choice)\\ & -35 = 1 \cdot (-35) && \text{and} && 1 + (-35) = -34\\ & -35 = -5 \cdot 7 && \text{and} && -5 + 7 = 2\\ & -35 = 5 \cdot (-7) && \text{and} && 5 + (-7) = -2\end{align*}

The answer is \begin{align*}(x - 1)(x + 35)\end{align*}.

### Review

1. \begin{align*}x^2 - 11x + 24\end{align*}
2. \begin{align*}x^2 - 13x + 42\end{align*}
3. \begin{align*}x^2 - 14x + 33\end{align*}
4. \begin{align*}x^2 - 9x + 20\end{align*}
5. \begin{align*}x^2 + 5x - 14\end{align*}
6. \begin{align*}x^2 + 6x - 27\end{align*}
7. \begin{align*}x^2 + 7x - 78\end{align*}
8. \begin{align*}x^2 + 4x - 32\end{align*}
9. \begin{align*}x^2 - 12x - 45\end{align*}
10. \begin{align*}x^2 - 5x - 50\end{align*}
11. \begin{align*}x^2 - 3x - 40\end{align*}
12. \begin{align*}x^2 - x - 56\end{align*}
13. \begin{align*}-x^2 - 2x - 1\end{align*}
14. \begin{align*}-x^2 - 5x + 24\end{align*}
15. \begin{align*}-x^2 + 18x - 72\end{align*}
16. \begin{align*}-x^2 + 25x - 150\end{align*}
17. \begin{align*}x^2 + 21x + 108\end{align*}
18. \begin{align*}-x^2 + 11x - 30\end{align*}
19. \begin{align*}x^2 + 12x - 64\end{align*}
20. \begin{align*}x^2 - 17x - 60\end{align*}
21. \begin{align*}x^2 + 5x - 36\end{align*}

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### Vocabulary Language: English

A quadratic polynomial is a polynomial of the 2nd degree, in other words, a polynomial with an $x^2$ term.