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

Factoring \(ax^2 + bx + c\) where \(a \ne 0, 1\)


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Types of Factoring

 

Factoring \(ax^2+bx+c\): Inspection Steps

Steps to Factor \(ax^2+bx+c\) by Inspection

 

Factoring \(6x^2-11x+4\): By Inspection Step 1

Factor \(6x^2-11x+4\).

Step 1: Remove any common factors

 

Factoring \(6x^2-11x+4\): By Inspection Step 2

Factor \(6x^2-11x+4\).

Step 2a: Determine binomial pairs that when expanded produce \(6x^2\)

 

Factoring \(6x^2-11x+4\): By Inspection Step 3

Factor \(6x^2-11x+4\).

Step 3: Eliminate options with common factors

 

Factoring \(6x^2-11x+4\): By Inspection Step 4

Factor \(6x^2-11x+4\).

Step 4: Expand and simplify to check

Four remaining candidates:

Check the first candidate:

 \(\begin{align*} (x+4)(6x+1) ~&\class{timed in1}{= 6x^2+x+24x+4}\\ &\class{timed in1}{=6x^2+25x+4} \end{align*}\)   

 

Factoring \(ax^2+bx+c\): Method of Decomposition Steps

Method of Decomposition

 

Factoring \(6x^2-11x+4\): By Decomposition Step 1 and 2

Factor \(6x^2-11x+4\).

 

 

Factoring \(6x^2-11x+4\): By Decomposition Step 3 and 4

Factor \(6x^2-11x+4\).

Solution — Decomposition Method 

Step 3: Decompose the middle term using the integers from Step 2

Decompose the middle term using \(-3\) and \(-8\):

 

 

Factoring \(6x^2-11x+4\): By Decomposition Step 5

Factor \(6x^2-11x+4\).

Solution — Decomposition Method 

Step 5: Common factor the bracket out of the two remaining terms

\(6x^2-11x+4\)

\(=6x^2-3x-8x+4\)

 

\(=3x\class{timed hl2}{(2x-1)} -4 \class{timed hl2}{(2x-1)}\)

 

 

 

Factoring \(6x^2-11x+4\): By Decomposition Alternate Method

Factor \(6x^2-11x+4\).

Alternate Solution

 

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Check Your Understanding 7


Factor \(((((((((s)*(tan(d)))*(a))*(r))*(d))*(F))*(o))*(r))*(m)\).

Enter factored form as "(ax+b)(cx+d)".

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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Example 7

Factor \(40x^2-14x-12\).

Solution

Before applying the Method of Decomposition, notice that there is a common factor of \(2\):

\(40x^2-14x-12 =2(20x^2-7x-6)\)

Now, we apply the Method of Decomposition to \(20x^2-7x-6\).

We require a pair of integers, with the following properties.

  • Add to \(-7\)
  • Multiply to \(20\times-6=-120\)

The two integers are \(-15\) and \(8\).

Thus,

\(\begin{align*} 40x^2-14x-12 & =2(20x^2-7x-6)\\ &=2(20x^2-15x+8x-6)\\ &=2\Big[5x(4x-3)+2(4x-3)\Big]\\ &=2(4x-3)(5x+2) \end{align*}\)


Check Your Understanding 8


Factor \(((((((((s)*(tan(d)))*(a))*(r))*(d))*(F))*(o))*(r))*(m)\).

Enter in factored form as "a(bx+c)(dx+e)".

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:

null

 

How Did I Do?Try Another