2. Unit 3: Algebraic Skills
  3. Lesson 2: Factoring — Common and Trinomials

Common Factoring


Four Types of Factoring

In this unit, we will look at four types of factoring:

  1. Common Factoring
  2. Factoring Trinomials of the Form \(ax^2+bx+c\) 
  3. Factoring a Difference of Squares 
  4. Factoring a Perfect Square

In this part, we will learn about common factoring.

Common Factoring

Expanding and factoring are opposites. For example,

  • \(5x^2(2x^3-3)\) expands to \(10x^5-15x^2\), and
  • \(10x^5-15x^2\) factors to \(5x^2(2x^3-3)\).

Recall

Expand
The algebra word meaning "multiply". Multiply until the expression is a collection of terms without brackets.
Factor
Write an expression as a product. Change an expression from a collection of two or more terms without brackets into a product with terms in brackets.

The first type of factoring is called common factoring. The name reminds us that we need to find what is in common to each term.

Consider \( 18x^3-30x^2\).

What factor is in common to the terms \(18x^3\) and \(-30x^2\)?

  • \(18\) and \(-30\) have a greatest common factor of \(6\). (Note that \(2\) and \(3\) are also common factors, but we require the greatest common factor).
  • \(x^3\) and \(x^2\) have a greatest common factor of \(x^2\). (Both terms have at least two \(x\)s multiplying together).

Combining these, the greatest factor in common to both \(18x^3\) and \(-30x^2\) is \(6x^2\).

Since \(6x^2\) is a factor of \(18x^3-30x^2\), we can find an expression whose product with \(6x^2\) is equal to \(18x^3-30x^2\). 

We need to determine this expression. In other words, we need to determine \(\diamondsuit\) and \(\triangle\) such that 

\[6x^2 (\diamondsuit + \triangle) = 18x^3-30x^2\]
  • There must be two terms inside the brackets since the expanded version has two terms.
  • The first term inside the brackets must be \(3x\) since \((6x^2)(3x)=18x^3\).
  • The second term inside the brackets must be \(-5\) since \((6x^2)(-5)=-30x^2 \).

Thus, the factored form of \(18x^3-30x^2\) is \(6x^2(3x-5)\).

Example 1

Factor fully.

  1. \(16x^2-8x+20\)
  2. \(45x^3y^6z^3+15xy^2\)

Solution

  1. \(16x^2-8x+20\)

    • The greatest common factor of \(16\), \(-8\), and \(20\) is \(4\).
    • There is not an \(x\) in common to all three terms (the last term has no \(x\)s).

    Thus, the greatest common factor of \(16x^2-8x+20\) is \(4\). 

    Next, we need to find the factor that multiplies with \(4\) to give \(16x^2-8x+20\).

    • The factor must have three terms since the expanded version has three terms.
    • Each term in the factor must multiply with \(4\) to equal the corresponding term in \(16x^2-8x+20\). 

    Thus,

    \(16x^2-8x+20 = 4(4x^2-2x+5)\)

  2. \(45x^3y^6z^3+15xy^2\)

    • The greatest common factor of \(45\) and \(15\) is \(15\).
    • There is an \(x\) in common to both terms.
    • There is a \(y^2\) in common to both terms.

    Thus, the greatest common factor of \(45x^3y^6z^3+15xy^2\) is \(15xy^2\). 

    Next, we need to find the factor that multiplies with \(15xy^2\) to give \(45x^3y^6z^3+15xy^2\).

    • The expression must have two terms since the expanded version has two terms.
    • Each term in the expression must multiply with \(15xy^2\) to equal the corresponding term in \(45x^3y^6z^3+15xy^2\).

    Thus,

    \(45x^3y^6z^3+15xy^2 = 15xy^2(3x^2y^4z^3+1)\)

Checking Our Work

We can check our answer by expanding.

For example, we can do this for part b) as follows:

\(\begin{align*}15xy^2(3x^2y^4z^3+1) & = (15xy^2)(3x^2y^4z^3) + (15xy^2) (1) \\ & =45x^3y^6z^3+15xy^2 \end{align*}\)


Check Your Understanding 1


Determine the greatest common factor of \(((((((((p)*(o))*(l))*(y))*(exp(a)))*(n))*(d))*(e))*(d)\).

Enter \(ax^by^c\) as "ax^by^c".

The greatest common factor is   There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:

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How Did I Do?Try Another

Check Your Understanding 2


Factor fully.    

\($fullExp(details...)\)

Enter \(3x^2(2x^2-1)\) in the form "3x^2(2x^2-1)".

Factored form: There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:

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How Did I Do?Try Another

Check Your Understanding 3


Factor fully.

\(((((((((p)*(o))*(l))*(y))*(exp(a)))*(n))*(d))*(e))*(d)\)

Enter \(3x^2y^3(2x^2-1)\) in the form "3x^2y^3(2x^2-1)".

Factored form: There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:

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How Did I Do?Try Another

Example 2

A golf ball is hit and the motion of the ball is described by \(h=-5t^2+30t\) where \(h\) is the ball's height above the ground (in metres) and \(t\) is the time (in seconds). Determine how long the ball is in the air.

Solution

To determine how long the ball is in the air, we need to know the times at which the ball's height is zero. To find the zeros, it is easier to work in factored form.

To factor \(h=-5t^2+30t\), notice the following:

  • The greatest common factor of \(-5\) and \(30\) is \(5\).
  • There is a \(t\) in common to both terms (each term has at least one \(t\)).

  • Since the leading coefficient is negative, we choose to factor out a \(-5\) not a \(5\).

Factoring the given expression:

\(\begin{align*} h & =-5t^2+30t\\ &= -5t(t-6) \end{align*}\)

Notice that since we factored out a negative \(5\) the sign inside the brackets changed since \((-5t)(-6)=+30t\) as required.

Using the factored form equation, we observe that:

  • The zeros are \(0\) and \(6\).
  • The ball starts on the ground at \(t=0\) and is on the ground again at \(t=6\).

Therefore, the ball is in the air for \(6\) seconds.

The graph of this relation confirms our results.

 

The graph has a max at x equals 3 at about 45 and has roots identified at x equal 0 and x equal 6. The parabola opens downward.

 

Summary

In common factoring, 

  • we remove the term that is "in common" to all terms, and
  • the result is a monomial multiplied by a polynomial.