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:
Factor the following trinomial with two variables: 6z2 + 11z + 4.
6z2 + 11z + 4 ⟹ 6z2 + 3z + 8z + 4
⟹ (6z2 + 3z) + (8z + 4)
⟹ 3z (2z + 1) + 4(2z + 1)
= (2z + 1) (3z + 4)
Factor 4a2 – 4ab + b2
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)
Factor x4 – 10x2y2 + 25y4
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)
Factor 2x2 + 7xy − 15y2
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)
Factor 4a7b3 – 10a6b2 – 24a5b.
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).
Factor 2a³ – 3a²b + 2a²c
Factor out the GCF, which a2
2a³ – 3a²b + 2a²c ⟹ a2(2a -3b + 2c)
Factor 9x² – 24xy + 16y²
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)
Factor pq – pr – 3ps
p is the common factor all the terms, therefore factor it out;
pq – pr – 3ps ⟹ p (q – r- 3s)