What is the process of multiplying trinomials?
By Forinfos - 15/02/2026 - 0 comments
Multiplying trinomials can seem like a daunting task. The complexity can be reduced by breaking the problem down and multiplying each term separately. In this example, the trinomials x^3 + 4x^2 + 3x + 7 and 8x^3 + 6x^2 + 5x + 2 are multiplied.
- Multiply the first term of the first trinomial with every term in the second trinomial
Take x^3 and multiply it by every term in the second trinomial, 8x^3 + 6x^2 + 5x + 2. The formula is x^3(8x^3) + x^3(6x^2) + x^3(5x) + x^3(2). The answer is 8x^6 + 6x^5 + 5x^4 + 2x^3 and is used in a later step.
- Multiply the second term of the first trinomial with every term in the second trinomial
Take 4x^2 and multiply it by every term in the second trinomial, 8x^3 + 6x^2 + 5x + 2. The formula is 4x^2(8x^3) + 4x^2(6x^2) + 4x^2(5x) + 4x^2(2). The answer is 16x^5 + 8x^4 + 20x^3 + 8x^2 and is used in a later step.
- Multiply the third term of the first trinomial with every term in the second trinomial
Take 3x and multiply it by every term in the second trinomial, 8x^3 + 6x^2 + 5x + 2. The formula is 3x(8x^3) + 3x(6x^2) + 3x(5x) + 3x(2). The answer is 24x^4 + 18x^3 + 15 x^2 + 6x and is used in a later step.
- Multiply the fourth term of the first trinomial with every term in the second trinomial
Take 7 and multiply it by every term in the second trinomial, 8x^3 + 6x^2 + 5x + 2. The formula is 7(8x^3) + 7(6x^2) + 7(5x) + 7(2), which reduces to 56x^3 + 42x^2 + 35x + 14. This answer, along with the answers from the previous steps, is used in the next step.
- Add the answers from the previous steps together
Take the answers from the previous steps, 8x^6 + 6x^5 + 5x^4 + 2x^3, 16x^5 + 8x^4 + 20x^3 + 8x^2, 24x^4 + 18x^3 + 15x^2 + 6x and 56x^3 + 42x^2 + 35x + 14, and add the like terms together. Reduce the equation to 8x^6 + (6x^5 + 16x^5) + (5x^4 + 8x^4 + 24x^4) + (2x^3 + 20x^3 + 18x^3 + 56x^3) + (8x^2 + 15x^2 + 42x^2) + (6x + 35x) + 14 to get the final answer of 8x^6 + 22x^5 + 37x^4 + 96x^3 + 65x^2 + 41x + 14.

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