Solving trinomial equations is a matter of determining the degree of the trinomial factoring according to the degree and the terms themselves and by then verifying that the factors found are the actual roots of the equation. These roots are the solutions to the equation.
Continue ReadingIn any trinomial expression, each individual term will have a power of x (x, x^2, x^3, ...). The highest power of these three terms determines the degree of the trinomial. If there are only powers of 1 for x, it is linear. If the highest power is 2, then it is quadratic. A power of 3 within the expression is called cubic, and this list can continue indefinitely. There are few examples of trinomials with powers of 4 or higher that are completely solvable, so it is best to show what can be solved.
If the equation turns out to be linear, it is solvable with algebra to combine constants or terms with x and get to a form of mx + n = 0 or y = mx + n, where m and n are real numbers. There will be one root/solution to the equation.
For quadratic trinomials of the form ax^2+bx+c, many equations are solved by factoring. Start with factoring c down to its prime numbers and determine all possible multiples between those prime numbers resulting in c. For example, c=6 will yield 2x3 as the prime numbers, but the multiples will be 6x1 or 2x3. When c is negative, seek one of these pairings where one number being negative and one number being positive will add to b. When c is positive, seek a pairing that adds to b. Only when c is negative will you need to worry about getting the correct sign. When c is positive, the sign of b determines the signs of the factors. Ex.: x^2-5x-14 = 0 First, 14 is factorable as 7x2 or 14x1. Since c is negative, one positive (2) and one negative (7) in the pairing (2,7) will add to -5. Thus, x^2-5x-14 = (x+2)(x-7) = 0. Remember that products that multiply to 0 indicate that one of the terms is 0. So x+2 = 0 and x-7 = 0. Then the roots/solutions are x=-2 and x=7. The quadratic equation also can solve these trinomials. The formula is [-b ± ? (b^2 - 4ac)] / 2a.
Not every cubic trinomial is solvable, but if your trinomial becomes a sum or difference of cubes, it is solvable by : x^3+a^3 = (x+a)(x^2-ax+a^2) x^3-a^3 = (x-a)(x^2+ax+a^2) In most other cases, there will be two or more terms that have some power of x in them. When this is the case, factor the highest common power of x from those terms and seek a solution to the remaining polynomial that remains. Ex: x^3 + 6x^2 - 4 = 0 Factoring x^2 from the first 2 terms, the trinomial becomes (x^2)(x+6) - 4 = 0, which then becomes (x^2)(x+6) = 4, yielding x = ±2.