2. Unit 3: Algebraic Skills
  3. Lesson 3: Factoring — Difference of Squares and Perfect Squares

Factoring Perfect Squares


A Return to \(ax^2+bx+c\)

Recall

  • To factor \(x^2+bx+c\), use the fact that \(x^2+bx+c=(x+p)(x+q)\) where \(p\) and \(q\) add to \(b\) and multiply to \(c\).

    For example,

    \[x^2-5x-66 = (x-11)(x+6)\]

    The two integers with sum \(-5\) and product \(-66\) are \(-11\) and \(6\).

  • To factor \(ax^2+bx+c\) when \(a \ne 1\), use the Method of Decomposition:
    1. If possible, common factor first.
    2. Think of two integers, \(p\) and \(q\), that add to \(b\) and multiply to \(ac\).
    3. Decompose the middle term using the integers from Step 2.
    4. Common factor the first two terms and second two terms. Ensure the resulting two brackets contain the same expressions.
    5. Common factor the common bracket out of the two remaining terms.

    For example,

    \[\begin{align*} 10x^2+11x-6& =10x^2+15x-4x-6\\ & =5x(2x+3)-2(2x+3)\\ &=(2x+3)(5x-2) \end{align*}\]

    The two integers with sum \(11\) and product \(-60\) are \(15\) and \(-4\).

Example 3

Apply the Method of Decomposition to factor these expressions.

  1. \(25x^2+40x+16\)
  2. \(4p^6-12p^3q^2+9q^4\)

Solution — Part A

\(25x^2+40x+16\)

We require two integers that

  • add to \(40\), and
  • multiply to \((25)(16)=400\).

The integers are \(20\) and \(20\).

Apply Decomposition,

\(25x^2+40x+16\\ =25x^2+20x+20x+16\\ =5x(5x+4)+4(5x+4)\\ =(5x+4)(5x+4)\\ =(5x+4)^2 \)

Solution — Part B

Recall part b): Apply the Method of Decomposition to factor \(4p^6-12p^3q^2+9q^4\).

\(4p^6-12p^3q^2+9q^4\)

This does not immediately look like \(ax^2+bx+c\) for two reasons:

It turns out that, given the right coefficients, Decomposition can be applied to \(ax^2+bxy+cy^2\).

In this case, if \(x=p^3\) and \(y=q^2\), then \(4p^6-12p^3q^2+9q^4=4x^2-12xy+9y^2\), which is of the form \(ax^2+bxy+cy^2\). 

Thus, we proceed with Decomposition.

We require two integers that 

  • add to \(-12\), and
  • multiply to \((4)(9)=36\).

The two integers are \(-6\) and \(-6\).

Applying Decomposition,

\(4p^6-12p^3q^2+9q^4\\ =4p^6-6p^3q^2-6p^3q^2+9q^4\\ =2p^3(2p^3-3q^2)-3q^2(2p^3-3q^2)\\ =(2p^3-3q^2)(2p^3-3q^2)\\ =(2p^3-3q^2)^2 \)

  • The exponents on \(p\) in the given expression are \(6\) and \(3\), not \(2\) and \(1\).
  • The given expression has two variables, whereas \(ax^2+bx+c\) does not.

Notice that both of these examples are perfect squares.

That is, when the factored form is found, the two factors are the same. 

Perfect Squares

How can we recognize a perfect square before factoring?

We can determine the pattern by expanding a general perfect square given in factored form:

\(\begin{align*}(a+b)^2& =(a+b)(a+b)\\ & =a^2+ab+ab+b^2\\ & =a^2+2ab+b^2 \end{align*}\)

Perfect Square: \(a^2+2ab+b^2=(a+b)^2\)

Previously, we showed that 

\[25x^2+40x+16 =(5x+4)^2\]

We can now see this from the formula using \(a=5x\) and \(b=4\):

  • \((a+b)^2\) is \((5x+4)^2\).
  • \(a^2+2ab+b^2\) is   \((5x)^2+2(5x)(4)+(4)^2\).
    • This simplifies to \(25x^2+40x+16\) as expected.

Example 4

Factor \(36x^8y^2-60x^4y+25\).

Solution

Method 1: Recognize as a perfect square

\(36x^8y^2-60x^4y+25\) is of the form \(a^2+2ab+b^2\) with \(a=6x^4y\) and \(b=-5\) since

  • \(a^2=(6x^4y)^2=36x^8y^2\)
  • \(2ab=2(6x^4y)(-5)=-60x^4y\)
  • \(b^2=(-5)^2=25\)

Also, \(a^2+2ab+b^2 = (a+b)^2\).

Thus, the factored form is \((a+b)^2\), which is \((6x^4y-5)^2\).

Method 2: Decomposition

We require two integers that 

  • add to \(-60\), and
  • multiply to \((36)(25)=900\).

The integers are \(-30\) and \(-30\).

Applying Decomposition,

\(\begin{align*} 36x^8y^2-60x^4y+25 &=36x^8y^2-30x^4y-30x^4y+25\\ &=6x^4y(6x^4y-5)-5(6x^4y-5)\\ &=(6x^4y-5)(6x^4y-5)\\ &=(6x^4y-5)^2 \end{align*}\)

Notice that both methods yield the same result, but the method of perfect squares lets us factor more efficiently.


Check Your Understanding 3


Factor \((((((((((((f)*(u))*(l))*(l))*(R))*(e))*(l))*(a))*(t))*(i))*(o))*(n)\).

Enter \(x^b\) as \(x\)^\(b\). Enter factors in parentheses.  Example: "(x^2-1)(x+3)".

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 5

Determine the zeros, axis of symmetry, and vertex of \(y=9x^2+ 42x +49\).

Solution

We proceed using three steps:

  1. Factor using either method (decomposition or recognize as a perfect square).
  2. Determine the zero(s).
  3. Use the zero(s) to determine the vertex.

Step 1: Factor using either method

Method 1: Decomposition

  • We require two integers that add to \(42\) and multiply to \((9)(49)=441\).
  • The integers are \(21\) and \(21\).

Applying the Method of Decomposition,

\(\begin{align*} y &= 9x^2+ 42x +49\\ &= 9x^2+21x+21x+49\\ &= 3x(3x+7)+7(3x+7)\\ &= (3x+7)(3x+7)\\ &= (3x+7)^2 \end{align*} \)

Method 2: Recognize as a perfect square

  • \(9x^2+42x+49\) is of the form \(a^2+2ab+b^2\) with \(a=3x\) and \(b=7\) since
    • \(a^2=(3x)^2=9x^2\),
    • \(2ab=2(3x)(7)=42x\), and
    • \(b^2=(7)^2=49\).
  • Also \((a+b)^2=a^2+2ab+b^2\)

Thus, the factored form is \((a+b)^2\), which is \((3x+7)^2\).

Step 2: Determine the zero(s)

Since \(y=(3x+7)^2\) is in factored form, we have the following:

  • The zeros are the \(x\)-values that make either factor \(0\).
  • In this case, there is only one unique factor (since the factors are equal), so there is only one zero.
  • We determine the zero by setting the factor \(3x+7\) equal to \(0\) and isolating for \(x\).

\(\begin{align*} 3x+7 &= 0\\ 3x &= -7\\ x &= \frac{-7}{3} \end{align*}\)

Step 3: Determine the vertex

Finally, we determine the vertex.

  • Normally, at this stage we average the two zeros to determine the axis of symmetry.
  • In this case, there is only one zero.

    The function has a vertex at the point (negative 7 over 3, 0).

  • Since there is only one zero, the vertex is on the \(x\)-axis.

Therefore,

  • the zero is \(\dfrac{-7}{3}\),
  • the axis of symmetry is \(x=\dfrac{-7}{3}\), and
  • the vertex is \(\left(\dfrac{-7}{3},0\right)\).

Check Your Understanding 4


Determine the vertex by first finding the zero(s) and axis of symmetry of \(y=$fullExp(details...)\). 

Answer must be exact. Enter \(\dfrac{a}{b}\) as "\(a / b\)". Enter the vertex as an ordered pair.  Example (1/2,3)

Vertex 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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