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Factor Trinomials Calculator

problem

Factor trinomial

$$ \color{blue}{ 4x^2-12x+9 } $$

solution

The factored form is:

$$ \color{blue}{ 4x^2-12x+9 = \left(2x-3\right)^2 } $$

explanation

Both the first and third terms are perfect squares.

$$ 4x^2 = \left( \color{blue}{ 2x } \right)^2 ~~ \text{and} ~~ 9 = \left( \color{red}{ 3 } \right)^2 $$

The middle term ( $ -12x $ ) is two times the product of the terms that are squared.

$$ -12x = - 2 \cdot \color{blue}{2x} \cdot \color{red}{3} $$

We can conclude that the polynomial $ 4x^{2}-12x+9 $ is a perfect square trinomial, so we will use the formula below.

$$ A^2 - 2AB + B^2 = (A - B)^2 $$

In this example we have $ \color{blue}{ A = 2x } $ and $ \color{red}{ B = 3 } $ so,

$$ 4x^{2}-12x+9 = ( \color{blue}{ 2x } - \color{red}{ 3 } )^2 $$

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Script name : factoring-trinomials-calculator

Form values: 4 , 2 , 12 , 1 , 9 , g , Factor tinomial 4x^2-12x+9 = 0 , Factor 4x^2-12x+9

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TUTORIAL

Polynomial Factoring Techniques

This calculator factors trinomials of the form $ ax^2 + bx + c $ using the methods listed below.

1. Factoring perfect square trinomial

2. Factor if leading coefficient $ a = 1 $

3. Factor if leading coefficient $ a \ne 1 $

4. Special cases ( $ b = 0 $ ) or ( $ a = 0 $ )

Method 1 : Factoring perfect square trinomial

Example 01: Factor $ 4a^2 - 12a + 9 $

Step1: Verify that both the first and third terms are perfect squares.

$4a^2$ is perfect square because $4a^2 = \left(\color{blue}{2a}\right)^2$

$9$ is perfect square because $9 = \left(\color{red}{3}\right)^2 $

Step2: Check if middle term is twice the product of $ \color{blue}{2a} $ and $ \color{red}{3} $

$$ \text{middle term} = 12a = 2 \cdot \color{blue}{2a} \cdot \color{red}{3} $$

Step3: Put $ \color{blue}{2a} $ and $ \color{red}{3}$ inside parentheses. Because the middle term's coefficient is negative, we'll insert a minus sign inside parenthesis.

$$ 4a^2 - 12a + 9 = ( \color{blue}{2a} - \color{red}{3} )^2 $$

Method 2 : Leading coefficient $ a = 1 $

In this case, the trinomial has the following form $ x^2 + bx + c $.

Example 02: Factor $ x^2 + 7x + 10 $

To factor this trinomial we need to find two integers ( $p$ and $q$ ) such that $ p + q = b $ and $ p \cdot q = c $.

In this example $ p + q = 7 $ and $p \cdot q = 10$

After some trials and errors we get $ p = 2 $ and $ q = 5 $

The factored form is

$$ x^2 + 7x + 10 = ( x + p)(x + q) = (x + 2)(x + 5) $$

Method 4 : Special Cases

Example 04: Factor $ 3x^2 + 5x $

This is special case where $c = 0$.

To solve this one we just need to factor $x$ out of $ 3x^2 + 5x $

$$ 3x^2 + 5x = x ( 3x + 5) $$

Example 05: Factor $ 25x^2 - 4 $

This is special case where $b = 0$.

We'll need to use the difference of squares formula to factor this one.

$$ 25x^2 - 4 = (5x)^2 - 2^2 = (5x-2)(5x+2) $$

Method 3 : Leading coefficient $ a \ne 1 $

In this case, the trinomial has the following form: $ ax^2 + bx + c $.

Example 03: Factor $ 3x^2 - 5x + 2 $

Step 1: Identify constants $a$ , $ b $ and $c$

$$ a = 3, b = -5 , c = 2 $$

Step 2: Find out two numbers ( $p$ and $q$) that multiply to $ a \cdot c = 6 $ and add up to $ b = -5 $.

After some trials and errors we get $ \color{blue}{p = -2} $ and $ \color{red}{q =-3} $

Step 3:Replace middle term ( $ -5x $ ) with $ \color{blue}{-2}x \color{red}{-3}x $

$$ 3x^2 - 5x + 2 = 3x^2 - 2x - 3x + 2 $$

Step 4:Factor out x from the first two terms and -1 from the last two terms.

$$ \begin{aligned} 3x^2 - 5x + 2 &= 3x^2 - 2x - 3x + 2 = \\\\ &= x(3x-2) -1(3x-2) = \\\\ &=(x - 1)(3x-2) \end{aligned} $$
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