Good question! A quadratic equation is an equation in which the variable is raised to the second power. Then, by remembering the zero product property, we know the solutions of the equation are the solutions of linear equations. Really, it’s another shining moment for the distributive property of multiplication! We’re not using it to calculate the product, but we can use it to write an expression as a product.įactoring is often used when solving quadratic equations because, then, the quadratic expression is rewritten as a product of linear expressions. The second pair would give us the factorization ( 2 x + 1 ) ( 3 x − 5 ) (2x+1)(3x-5) ( 2 x + 1 ) ( 3 x − 5 ), with inner and outer products 1 ⋅ 3 = 3 1\cdot3=3 1 ⋅ 3 = 3 and 2 ⋅ − 5 = − 10 2\cdot-5=-10 2 ⋅ − 5 = − 10.Just because math is getting more complicated doesn’t mean we can’t make it a little easier for ourselves! Factoring is one way we can make our calculations a little clearer and easier to follow.Īnd who wouldn’t want to make math easier? What does it mean to factor an expression?įactoring an expression means rewriting it as a product of two or more factors. Still not valid, but we're just a sign away. of each side, factoring, and completing the square. Then it will attempt to solve the equation by using one or more of the following: addition, subtraction, division, taking the square root. Let's consider the same pair of numbers that factorize the last coefficient:įor the first pair, we'd create the factorization ( 2 x − 1 ) ( 3 x + 5 ) (2x-1)(3x+5) ( 2 x − 1 ) ( 3 x + 5 ). When you enter an equation into the calculator, the calculator will begin by expanding (simplifying) the problem. Is this it? Nope!Ĭhange the choice of factors for the first coefficient: this time, we will use 2 2 2 and 3 3 3. Repeat steps 3 and 4 for the other pair of coefficients: the products will be 1 ⋅ 6 = 6 1\cdot6=6 1 ⋅ 6 = 6 and 1 ⋅ − 5 = − 5 1\cdot -5 = -5 1 ⋅ − 5 = − 5, which sum to 6 − 5 = 1 6-5=1 6 − 5 = 1: again, a nonvalid factorization. Let's compute the products of the inner and outer coefficients: − 1 ⋅ 6 = − 6 -1\cdot 6 = -6 − 1 ⋅ 6 = − 6 and 1 ⋅ 5 = 5 1\cdot 5 = 5 1 ⋅ 5 = 5.Ĭalculate the sum of these results: − 6 + 5 = − 1 -6+5=-1 − 6 + 5 = − 1. With such combinations, we'd create the factorization ( x − 1 ) ( 6 x + 5 ) (x-1)(6x+5) ( x − 1 ) ( 6 x + 5 ). Move onto the last coefficient: − 5 -5 − 5 (the sign matters): in this case, we have two possible choices: − 1 -1 − 1 and 5 5 5, and 1 1 1 and − 5 -5 − 5. Now that you know what the reverse FOIL is and how to calculate it, we can try to apply it to a polynomial with numerical coefficients! Let's take the trinomial 6 x 2 − 7 x − 5 6x^2-7x-5 6 x 2 − 7 x − 5.įind a pair of numbers that, multiplied, returns 6 6 6 (the first coefficient, a a a): 1 1 1 and 6 6 6 are a valid choices. If, even after exhausting all possible factorizations of a a a, you didn't find a valid combination of inner and outer products, it means you stumbled upon an irreducible polynomial! If this is the case, return to step 1, find another possible combination of α \alpha α and γ \gamma γ, and repeat the algorithm's steps from there. It may happen that no products allow you to find a valid factorization. If the previous step didn't return a valid factorization, repeat it for the next pair of products you found in step 3. Write the result as ( α x + β ) ( γ x + δ ) (\alpha x+\beta)(\gamma x + \delta) ( αx + β ) ( γ x + δ ). If the result equals b b b ( α ⋅ δ + β ⋅ γ = b \alpha\cdot\delta+\beta\cdot\gamma = b α ⋅ δ + β ⋅ γ = b), you found a factorization of the trinomial. If you want to calculate the reverse FOIL method, follow these steps:įind a pair of numbers α \alpha α and γ \gamma γ that satisfy α ⋅ γ = a \alpha\cdot\gamma = a α ⋅ γ = a.įind all possible pairs of numbers β \beta β and δ \delta δ such that β ⋅ δ = c \beta\cdot\delta=c β ⋅ δ = c.Ĭompute the products α ⋅ δ \alpha\cdot\delta α ⋅ δ (the product of the outer coefficients) and γ ⋅ δ \gamma\cdot\delta γ ⋅ δ (the product of the inner coefficients) for all possible pairs of β \beta β and δ \delta δ.Ĭompute the sum of a pair of products you found in the step above.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |