Factoring Trinomial with Two Variables – Method & Examples

A trinomial is an algebraic equation composed of three terms and is normally of the form ax² + 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; 2x² + 7xy − 15y², e² − 6ef + 9f², 2c² + 13cd + 6d², 30x³y – 25x²y² – 30xy³, 6x² – 17xy + 10y²etc.

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: 6z² + 11z + 4.

Solution

6z² + 11z + 4 ⟹ 6z² + 3z + 8z + 4

⟹ (6z² + 3z) + (8z + 4)

⟹ 3z (2z + 1) + 4(2z + 1)

= (2z + 1) (3z + 4)

Example 2

Factor 4a² – 4ab + b²

Solution

Apply the method of factoring a perfect square trinomial

4a² – 4ab + b² ⟹ (2a)² – (2)(2) ab + b²

= (2a – b)²

= (2a – b) (2a – b)

Example 3

Factor x⁴ – 10x²y² + 25y⁴

Solution

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

x⁴ – 10x²y² + 25y⁴ ⟹ (x²)² – 2 (x²) (5y²) + (5y²)²

Apply the formula a² + 2ab + b² = (a + b)² to get,

= (x² – 5y²)²

= (x² – 5y²) (x² – 5y²)

Example 4

Factor 2x² + 7xy − 15y²

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)

2x² + 7xy − 15y² ⟹2x² -3xy + 10xy − 15y²

Factor by grouping.

2x² -3xy + 10xy − 15y² ⟹x (2x – 3y) + 5y (2x -3y)

⟹ (x +5y) (2x -3y)

Example 5

Factor 4a⁷b³ – 10a⁶b² – 24a⁵b.

Solution

Factor out a 2a⁵b first.

4a⁷b³ – 10a⁶b² – 24a⁵b ⟹2a⁵b (2a²b² – 5ab – 12)

But since, 2a²b² – 5ab – 12 ⟹ (2x + 3) (x – 4)

Therefore, 4a⁷b³ – 10a⁶b² – 24a⁵b ⟹2a⁵b (2ab + 3) (ab – 4).

Example 6

Factor 2a³ – 3a²b + 2a²c

Solution

Factor out the GCF, which a²

2a³ – 3a²b + 2a²c ⟹ a²(2a -3b + 2c)

Example 7

Factor 9x² – 24xy + 16y²

Solution

Since both the first and last term are squared, then apply the formula a² + 2ab + b² = (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)