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# Factoring Trinomial with Two Variables â€“ 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}Â + 7xyÂ âˆ’Â 15y^{2}, e^{2Â }âˆ’Â 6ef +Â 9f^{2},Â 2c^{2}Â + 13cd + 6d^{2}, 30x^{3}y – 25x^{2}y^{2}Â – 30xy^{3}, 6x^{2}Â – 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.

*Letâ€™s solve a few examples of trinomials with two variables:*

*Example 1*

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

__Solution__

6z^{2}Â + 11zÂ + 4 âŸ¹ 6z^{2}Â + 3zÂ + 8zÂ + 4

âŸ¹ (6z^{2}Â + 3z) + (8zÂ + 4)

âŸ¹ 3z (2zÂ + 1) + 4(2zÂ + 1)

= (2zÂ + 1) (3zÂ + 4)

*Example 2*

FactorÂ 4a^{2}Â â€“ 4ab + b^{2}

__Solution__

Apply the method of factoring a perfect square trinomial

4a^{2}Â â€“ 4ab + b^{2} âŸ¹ (2a)^{2}Â â€“ (2)(2) ab + b^{2}

= (2a â€“ b)^{2}

= (2a â€“ b) (2a â€“ b)

*Example 3*

FactorÂ x^{4}Â – 10x^{2}y^{2}Â + 25y^{4}

__Solution__

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

x^{4} – 10x^{2}y^{2}Â + 25y^{4} âŸ¹ (x^{2})^{2}Â – 2 (x^{2}) (5y^{2}) + (5y^{2})^{2}

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

= (x^{2}Â – 5y^{2})^{2}

= (x^{2}Â â€“ 5y^{2}) (x^{2}Â â€“ 5y^{2})

*Example 4*

Factor 2x^{2}Â + 7xyÂ âˆ’Â 15y^{2}

__Solution__

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

âŸ¹Â 2*-15 = -30

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

âŸ¹ 10 * -3 = -30

âŸ¹ 10 + (-3) = 7

Therefore, the two numbers are -3 and 10.

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

2x^{2}Â + 7xyÂ âˆ’Â 15y^{2}Â âŸ¹2x^{2}Â -3xy + 10xy âˆ’Â 15y^{2}

Factor by grouping.

2x^{2}Â -3xy + 10xy âˆ’Â 15y^{2}Â âŸ¹x (2x â€“ 3y) + 5y (2x -3y)

âŸ¹ (x +5y) (2x -3y)

*Example 5*

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

__Solution__

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

4a^{7}b^{3Â }– 10a^{6}b^{2Â }– 24a^{5}b âŸ¹2a^{5}b (2a^{2}b^{2}Â – 5ab – 12)

But since, 2a^{2}b^{2}Â – 5ab â€“ 12 âŸ¹ (2x + 3) (x – 4)

Therefore, 4a^{7}b^{3Â }– 10a^{6}b^{2Â }– 24a^{5}b âŸ¹2a^{5}b (2ab + 3) (ab – 4).

*Example 6*

Factor 2aÂ³ – 3aÂ²b + 2aÂ²c

__Solution__

Factor out the GCF, which a^{2}

2aÂ³ – 3aÂ²b + 2aÂ²cÂ âŸ¹ a^{2}(2a -3b + 2c)

*Example 7*

FactorÂ 9xÂ² – 24xy + 16yÂ²

__Solution__

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

9xÂ² – 24xy + 16yÂ²Â âŸ¹3Â²Â xÂ² – 2(3x) (4y) + 4Â²Â yÂ²

âŸ¹ (3Â x) Â² – 2(3x) (4y) + (4Â y) Â²

âŸ¹Â (3xÂ – 4y) Â²

âŸ¹Â (3xÂ – 4y) (3xÂ – 4y)

*Example 8*

Factor pq – pr – 3ps

__Solution__

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

pq – pr – 3ps âŸ¹ p (q – r- 3s)