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Factoring when the Leading Coefficient Equals 1

Factor simple trinomials and difference of squares binomials

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Factoring Quadratics When the Lead Coefficient Equals 1

The area of a rectangle is \begin{align*}x^2-3x-28\end{align*}x23x28. What are the length and width of the rectangle?

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Watch the first few examples in this video, until about 12:40.

Khan Academy: Factoring Quadratic Expressions


A quadratic equation has the form \begin{align*}ax^2+bx+c\end{align*}ax2+bx+c, where \begin{align*}a \ne 0\end{align*}a0 (If \begin{align*}a = 0\end{align*}a=0, then the equation would be linear). For all quadratic equations, the 2 is the largest and only exponent. A quadratic equation can also be called a trinomial when all three terms are present.

There are four ways to solve a quadratic equation. In this concept, we are going to focus on factoring when the lead coefficient is one (i.e.,  \begin{align*}a = 1\end{align*}a=1).  In this case, there is no number in front of \begin{align*}x^2\end{align*}x2. First, let’s start with a review of multiplying two factors together.

Example A

Multiply \begin{align*}(x+4)(x-5)\end{align*}(x+4)(x5).

Solution: Even though this is not a quadratic, the product of the two factors will be. Remember from previous math classes that a factor is a number that goes evenly into a larger number. For example, 4 and 5 are factors of 20. So, to determine the larger number that \begin{align*}(x + 4)\end{align*}(x+4) and \begin{align*}(x-5)\end{align*}(x5) go into, we need to multiply them together.

One method for multiplying two polynomial factors together is called FOIL. To do this, you need to multiply the FIRST terms, OUTSIDE terms, INSIDE terms, and the LAST terms together and then combine like terms.  FOIL is the same as distributing each term in the first set of parentheses.  

In the example above, notice that FOIL is the same as distributing the x and then distributing the 4.

Therefore \begin{align*}(x+4)(x-5)=x^2-x-20\end{align*}(x+4)(x5)=x2x20. We can also say that \begin{align*}(x+4)\end{align*}(x+4) and \begin{align*}(x-5)\end{align*}(x5) are factors of \begin{align*}x^2-x-20\end{align*}x2x20.


Now, we will “undo” the multiplication of two factors by factoring. In this concept, we will only address quadratic equations in the form \begin{align*}x^2+bx+c\end{align*}x2+bx+c, or when \begin{align*}a = 1\end{align*}a=1.

Factor \begin{align*}x^2+6x+8\end{align*}x2+6x+8.



In the quadratic \begin{align*}x^2+bx+c\end{align*}x2+bx+c\begin{align*}b\end{align*}b will be equal to the sum of \begin{align*}m\end{align*}m and \begin{align*}n\end{align*}n and \begin{align*}c\end{align*}c will be equal to their product. Applying this to our problem, we ask "What two factors of 8 will add up to 6?"  

Write out a list of factor pairs for 8, if you need to:

License: CC BY-NC 3.0

The green pair above is the only one that also adds up to 6.  This means that 4 and 2 are the numbers we need.

\begin{align*}x^2+6x+8\\ (x+4)(x+2)\end{align*}x2+6x+8(x+4)(x+2)

Therefore, the factors of \begin{align*}x^2+6x+8\end{align*}x2+6x+8 are \begin{align*}(x+4)(x+2)\end{align*}(x+4)(x+2). You can FOIL this to check your answer.

Example C

Factor \begin{align*}x^2+12x-28\end{align*}x2+12x28.

Solution: We can approach this problem in exactly the same way we did Example B. What are the factors of -28 that add up to 12? Let’s list them out to see:

\begin{align*}-4 \cdot 7, 4 \cdot -7, 2 \cdot -14, {\color{red}-2 \cdot 14}, 1 \cdot -28, -1 \cdot 28\end{align*}47,47,214,214,128,128

The \begin{align*}{\color{red}\mathbf{red}}\end{align*} pair above is the one that works. Notice that we only listed the factors of negative 28.

\begin{align*}x^2+12x-28\\ (x-2)(x+14)\end{align*}

Example D

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

Solution: This is an example of a quadratic that is not a trinomial because it only has two terms, also called a binomial. There is no \begin{align*}c\end{align*}, or constant term. To factor this, we need to look for the GCF. In this case, the largest number that can be taken out of both terms is an \begin{align*}x\end{align*}.


Therefore, the factors are \begin{align*}x\end{align*} and \begin{align*}x-4\end{align*}.

Intro Problem Revisit Recall that the area of a rectangle is \begin{align*}A = lw\end{align*}, where l is the length and w is the width. To find the length and width, we can therefore factor the area \begin{align*}x^2-3x-28\end{align*}.

What are the factors of –28 that add up to –3? Testing the various possibilities, we find that \begin{align*}-7 \cdot 4 = -28\end{align*} and \begin{align*}-7 + 4 = -3\end{align*}.

Therefore, \begin{align*}x^2-3x-28\end{align*} factors to \begin{align*}(x - 7)(x + 4)\end{align*}, and one of these factors is the rectangle's length while the other is its width.

Guided Practice

1. Multiply \begin{align*}(x-3)(x + 8)\end{align*}.

Factor the following quadratic expressions. If it cannot be factored, write not factorable.

2. \begin{align*}x^2-9x+20\end{align*}

3. \begin{align*}x^2+7x-30\end{align*}

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

5. \begin{align*}x^2+10x\end{align*}


1. FOIL-ing our factors together, we get:

\begin{align*}(x-3)(x+8)= x^2+8x-3x-24 = x^2+5x-24\end{align*}

2. Ask yourself, "What two factors of 20 will add up to -9?  \begin{align*}-4 + -5 = -9\end{align*} and \begin{align*}-4 \cdot -5 = 20\end{align*}.


3. Let’s list out all the factors of -30 and their sums. The sums are in red.

\begin{align*}-10 \cdot 3 \ {\color{red}(-7)}, -3 \cdot 10 \ {\color{red}(7)}, -2 \cdot 15 \ {\color{red}(13)}, -15 \cdot 2 \ {\color{red}(-13)}, -1 \cdot 30 \ {\color{red}(29)}, -30 \cdot 1 \ {\color{red}(-29)}\end{align*}

From this, the factors of -30 that add up to 7 are -3 and 10. \begin{align*}x^2+7x-30 = (x-3)(x+10)\end{align*}

4. There are no factors of 6 that add up to 1. If we had -6, then the trinomial would be factorable. But, as is, this is not a factorable trinomial.

5. The only thing we can do here is to take out the GCF. \begin{align*}x^2+10x=x(x+10)\end{align*}


Quadratic Equation
An equation where the largest exponent is a 2 and has the form \begin{align*}ax^2+bx+c\end{align*}, \begin{align*}a \ne 0\end{align*}.
A quadratic equation with three terms.
A quadratic equation with two terms.
A way to break down an expression into smaller factors.
A number that goes evenly into a larger number.
A method used to multiply together two factors. You multiply the FIRST terms, OUTSIDE terms, INSIDE terms, and LAST terms and then combine any like terms.
The number in front of a variable.
A number that is added or subtracted within an equation.


Multiply the following factors together.

  1. \begin{align*}(x+2)(x-8)\end{align*}
  2. \begin{align*}(x-9)(x-1)\end{align*}
  3. \begin{align*}(x+7)(x+3)\end{align*}

Factor the following quadratic expressions. If it cannot be factored, write not factorable

  1. \begin{align*}x^2-x-2\end{align*}
  2. \begin{align*}x^2+2x-24\end{align*}
  3. \begin{align*}x^2-6x\end{align*}
  4. \begin{align*}x^2+6x+9\end{align*}
  5. \begin{align*}x^2+8x-10\end{align*}
  6. \begin{align*}x^2-11x+30\end{align*}
  7. \begin{align*}x^2+13x-30\end{align*}
  8. \begin{align*}x^2+11x+28\end{align*}
  9. \begin{align*}x^2-8x+12\end{align*}
  10. \begin{align*}x^2-7x-44\end{align*}
  11. \begin{align*}x^2-8x-20\end{align*}
  12. \begin{align*}x^2+4x+3\end{align*}
  13. \begin{align*}x^2-5x+36\end{align*}
  14. \begin{align*}x^2-5x-36\end{align*}
  15. \begin{align*}x^2+x\end{align*}

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