The Perfect Square Trinomial

What is a perfect square trinomial?

Definition

A perfect square trinomial is a special form of quadratic trinomial that can be factored into a binomial squared ($(ax + b)^2$ or $(ax - b)^2$). It arises when you square a binomial, resulting in a trinomial where:

1. The first term is a perfect square.
2. The third term is also a perfect square.
3. The middle term is twice the product of the square roots of the first and third terms.

Characteristics and Formula:

A perfect square trinomial has the general form: $a^2x^2 + 2abx + b^2$ (when expanded from $(ax + b)^2$) or $a^2x^2 - 2abx + b^2$ (when expanded from $(ax - b)^2$)

Key Points:

• Square of the First Term: The first term of the trinomial comes from squaring the first term of the binomial.
• Square of the Last Term: The last term of the trinomial is the square of the second term of the binomial.
• Middle Term: The middle term is crucial for identification. It is twice the product of the first and second terms of the binomial.

How to Identify a Perfect Square Trinomial:

To check if a given trinomial $c^2x^2 + dx + e$ is a perfect square:

1. Ensure that both $c^2x^2$ and $e$ are perfect squares.
2. Check if the middle term $dx$ is equal to $2 \cdot c \cdot \sqrt{e}$ if $c^2x^2$ and $e$ are positive or equal to $-2 \cdot c \cdot \sqrt{e}$ if $c^2x^2$ and $e$ are negative. Note that $d$ should be the result of doubling the product of the square root of $c^2x^2$ (which is $cx$) and the square root of $e$.

Examples:

1. $x^2 + 6x + 9$: This is a perfect square trinomial because it can be factored into $(x + 3)^2$. Here, $x^2$ is the square of $x$, $9$ is the square of $3$, and $6x$ is twice the product of $x$ and $3$.

2. $4x^2 - 12x + 9$: It factors into $(2x - 3)^2$, making it a perfect square trinomial. $4x^2$ is the square of $2x$, $9$ is the square of $3$, and $-12x$ is twice the product of $2x$ and $-3$.

How to factor a perfect square trinomial?

Factoring a perfect square trinomial means expressing it in the form $(ax + b)^2$ or $(ax - b)^2$, where the original trinomial fits the pattern $a^2x^2 \pm 2abx + b^2$.

Step 1: Identify the Square Roots

• First Term: Find the square root of the coefficient of $x^2$, which gives you $a$.
• Third Term: Similarly, find the square root of the constant term, which provides $b$.

Step 2: Check the Middle Term

Confirm that the middle term (the coefficient of $x$) is exactly twice the product of the square roots from Step 1. If it is positive, you'll use addition in the binomial; if negative, you'll use subtraction.

Step 3: Construct the Binomial

• Write down the binomial as $(ax \pm b)^2$, where $\pm$ depends on the sign of the middle term in the trinomial.

Solved examples of factoring factor a perfect square trinomial.

Example 1: Basic Perfect Square

Given the trinomial $x^2 + 6x + 9$.

Step 1: Identify Square Roots

• First Term: The square root of $x^2$ is $x$.
• Third Term: The square root of $9$ is $3$.

Step 2: Confirm Middle Term

• Middle Term: The middle term, $6x$, is twice the product of $x$ and $3$ ($2 \times x \times 3 = 6x$).

Step 3: Factor

• The trinomial factors into: $(x + 3)^2$.

Example 2: Coefficient Greater Than 1

Consider the trinomial $4x^2 + 12x + 9$.

Step 1: Identify Square Roots

• First Term: The square root of $4x^2$ is $2x$.
• Third Term: The square root of $9$ is $3$.

Step 2: Confirm Middle Term

• Middle Term: $12x$ is twice the product of $2x$ and $3$ ($2 \times 2x \times 3 = 12x$).

Step 3: Factor

• The trinomial factors into: $(2x + 3)^2$.

Example 3: Negative Middle Term

Given $x^2 - 8x + 16$.

Step 1: Identify Square Roots

• First Term: The square root of $x^2$ is $x$.
• Third Term: The square root of $16$ is $4$.

Step 2: Confirm Middle Term

• Middle Term: $-8x$ corresponds to twice the product of $x$ and $-4$ ($2 \times x \times (-4) = -8x$).

Step 3: Factor

• Therefore, the trinomial factors into: $(x - 4)^2$.

Example 4: Larger Coefficients

Consider $9x^2 + 30x + 25$.

Step 1: Identify Square Roots

• First Term: The square root of $9x^2$ is $3x$.
• Third Term: The square root of $25$ is $5$.

Step 2: Confirm Middle Term

• Middle Term: $30x$ is twice the product of $3x$ and $5$ ($2 \times 3x \times 5 = 30x$).

Step 3: Factor

• The trinomial factors into: $(3x + 5)^2$.

Example 5: With Subtraction

Given $16y^2 - 40y + 25$.

Step 1: Identify Square Roots

• First Term: The square root of $16y^2$ is $4y$.
• Third Term: The square root of $25$ is $5$.

Step 2: Confirm Middle Term

• Middle Term: $-40y$ matches twice the product of $4y$ and $-5$ ($2 \times 4y \times (-5) = -40y$).

Step 3: Factor

• Hence, we get: $(4y - 5)^2$.

FAQs about the perfect square trinomial

Q: What is the formula for cubed trinomial?

A: For the Cube of a Sum $(a + b)^3$:

$ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $

For the Cube of a Difference $(a - b)^3$:

$ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $

Formula Breakdown:

• $a^3$: The cube of the first term.
• $3a^2b$: Three times the square of the first term multiplied by the second term.
• $3ab^2$: Three times the first term multiplied by the square of the second term.
• $b^3$: The cube of the second term.

Example:

To illustrate the application, let's find the cube of $(x + 2)^3$ using the formula:

$ (x + 2)^3 = x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3 $ $ = x^3 + 6x^2 + 12x + 8 $

And for the cube of a difference, $(x - 2)^3$:

$ (x - 2)^3 = x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 $ $ = x^3 - 6x^2 + 12x - 8 $

Q: Is the difference of two trinomials always a trinomial?

A: No, the difference of two trinomials is not always a trinomial. When you subtract one trinomial from another, the result can be any polynomial, depending on the specific terms of the trinomials involved and how they simplify.

Trinomials Definition:

A trinomial is a polynomial with three terms.

Example: Result is Not a Trinomial

However, consider a case where like terms cancel out:

• $A(x) = 3x^2 + 5x + 6$
• $B(x) = 3x^2 + 3x + 4$

Subtraction $A(x) - B(x)$ results in:

• $3x^2 + 5x + 6 - (3x^2 + 3x + 4) = (3x^2 - 3x^2) + (5x - 3x) + (6 - 4) = 2x + 2$

The result in this case is a binomial, not a trinomial.