- Calculators
- ::
- Quadratic equations and functions
- ::
- Factoring Trinomials
Trinomials Factoring Calculator
This calculator factors trinomials of the form $ax^2+bx+c$ using the AC method and a formula $ax^2+bx+c = a(x-x_1)(x-x_2)$, where $x_1$ and $x_2$ are solutions of a quadratic equation. Calculator shows all the work and provides detailed explanation for each step.
Factor trinomial $$ \color{blue}{ 3x^2-5x+2 } $$
solution
The factored form is $$ \color{blue}{ 3x^2-5x+2 = \left(x-1\right)\left(3x-2\right) } $$
explanation
Step 1: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:
Step 2: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = 2} $.
$$ a \cdot c = 6 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 6 $ and add to $ b = -5 $.
Step 4: All pairs of numbers with a product of $ 6 $ are:
| PRODUCT = 6 | |
| 1 6 | -1 -6 |
| 2 3 | -2 -3 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -5 }$
| PRODUCT = 6 and SUM = -5 | |
| 1 6 | -1 -6 |
| 2 3 | -2 -3 |
Step 6: Replace middle term $ -5 x $ with $ -2x-3x $:
$$ 3x^{2}-5x+2 = 3x^{2}-2x-3x+2 $$Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -1 $ out of the last two terms.
$$ 3x^{2}-2x-3x+2 = x\left(3x-2\right) -1\left(3x-2\right) = \left(x-1\right) \left(3x-2\right) $$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.
$$ \begin{aligned} 3x^2 - 5x + 2 &= 3x^2 - 2x - 3x + 2 = \\\\ &= x(3x-2) -1(3x-2) = \\\\ &=(x - 1)(3x-2) \end{aligned} $$Please tell me how can I make this better.