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# Factoring Trinomial with Two Variables – Method & Examples

A trinomial is an algebraic equation composed of three terms and is normally of the form ax^{2 }+ bx + c = 0, where a, b and c are numerical coefficients.

To **factor a trinomial is to decompose an equation into the product of two or more binomials**. This means that we will rewrite the trinomial in the form (x + m) (x + n).

## Factoring Trinomials with Two Variables

Sometimes, a trinomial expression may consist of only two variables. This trinomial is known as a bivariate trinomial.

Examples of bivariate trinomials are; 2x^{2} + 7xy − 15y^{2}, e^{2 }− 6ef + 9f^{2}, 2c^{2} + 13cd + 6d^{2}, 30x^{3}y – 25x^{2}y^{2} – 30xy^{3}, 6x^{2} – 17xy + 10y^{2}etc.

A trinomial with two variables are factored similarly as if it has only one variable.

**Different factoring methods** such as reverse FOIL method, perfect square factoring, factoring by grouping, and the AC method can solve these kinds of trinomials with two variables.

## How to Factor Trinomials with Two Variables?

*To factor a trinomial with two variables, the following steps are applied:*

- Multiply the leading coefficient by the last number.
- Find the sum of two numbers that add to the middle number.
- Split the middle term and group in twos by removing the GCF from each group.
- Now, write in factored form.

*Let’s solve a few examples of trinomials with two variables:*

*Example 1*

Factor the following trinomial with two variables: 6z^{2} + 11z + 4.

__Solution__

6z^{2} + 11z + 4 ⟹ 6z^{2} + 3z + 8z + 4

⟹ (6z^{2} + 3z) + (8z + 4)

⟹ 3z (2z + 1) + 4(2z + 1)

= (2z + 1) (3z + 4)

*Example 2*

Factor 4a^{2} – 4ab + b^{2}

__Solution__

Apply the method of factoring a perfect square trinomial

4a^{2} – 4ab + b^{2} ⟹ (2a)^{2} – (2)(2) ab + b^{2}

= (2a – b)^{2}

= (2a – b) (2a – b)

*Example 3*

Factor x^{4} – 10x^{2}y^{2} + 25y^{4}

__Solution__

This trinomial is a perfect, therefore apply the perfect square formula.

x^{4} – 10x^{2}y^{2} + 25y^{4} ⟹ (x^{2})^{2} – 2 (x^{2}) (5y^{2}) + (5y^{2})^{2}

Apply the formula a^{2} + 2ab + b^{2} = (a + b)^{2} to get,

= (x^{2} – 5y^{2})^{2}

= (x^{2} – 5y^{2}) (x^{2} – 5y^{2})

*Example 4*

Factor 2x^{2} + 7xy − 15y^{2}

__Solution__

Multiply the leading coefficient by the coefficient of the last term.

⟹ 2*-15 = -30

Find two numbers product is -30 and sum is 7.

⟹ 10 * -3 = -30

⟹ 10 + (-3) = 7

Therefore, the two numbers are -3 and 10.

Replace the middle term of the original trinomial with (-3xy +10xy)

2x^{2} + 7xy − 15y^{2} ⟹2x^{2} -3xy + 10xy − 15y^{2}

Factor by grouping.

2x^{2} -3xy + 10xy − 15y^{2} ⟹x (2x – 3y) + 5y (2x -3y)

⟹ (x +5y) (2x -3y)

*Example 5*

Factor 4a^{7}b^{3 }– 10a^{6}b^{2 }– 24a^{5}b.

__Solution__

Factor out a 2a^{5}b first.

4a^{7}b^{3 }– 10a^{6}b^{2 }– 24a^{5}b ⟹2a^{5}b (2a^{2}b^{2} – 5ab – 12)

But since, 2a^{2}b^{2} – 5ab – 12 ⟹ (2x + 3) (x – 4)

Therefore, 4a^{7}b^{3 }– 10a^{6}b^{2 }– 24a^{5}b ⟹2a^{5}b (2ab + 3) (ab – 4).

*Example 6*

Factor 2a³ – 3a²b + 2a²c

__Solution__

Factor out the GCF, which a^{2}

2a³ – 3a²b + 2a²c ⟹ a^{2}(2a -3b + 2c)

*Example 7*

Factor 9x² – 24xy + 16y²

__Solution__

Since both the first and last term are squared, then apply the formula a^{2} + 2ab + b^{2} = (a + b)^{2} to get,

9x² – 24xy + 16y² ⟹3² x² – 2(3x) (4y) + 4² y²

⟹ (3 x) ² – 2(3x) (4y) + (4 y) ²

⟹ (3x – 4y) ²

⟹ (3x – 4y) (3x – 4y)

*Example 8*

Factor pq – pr – 3ps

__Solution__

p is the common factor all the terms, therefore factor it out;

pq – pr – 3ps ⟹ p (q – r- 3s)