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Algebra 1A. Look, Look, Look, Method! MENTAL MATH. Mental Math Method for Factoring Second Degree Polynomials
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Algebra1A Look, Look, Look, Method! MENTAL MATH Izydorczak 2014
Mental Math Method for Factoring Second Degree Polynomials Sometimes referred to as reverse FOIL, this method for factoring a second degree polynomial requires you to mentally picture the two binomials that might be factors of the second degree polynomial. For example, suppose you are asked to factor x2 + 5x + 6. The thought process might go something like this: 1) The sign of the last term (6) is positive. Therefore the signs of the two binomial factors are the same (since multiplying like signs gives a positive number; if the sign had been negative, we know the binomial factors would have different signs). 2) The variable parts of the binomial factors are both going to be x, because only x * x (the multiplication of the "firsts") will give me x2. 3) I know the number parts of the binomials have to be multiply to be 6, since I will be multiplying them together to get the last term of the polynomial. Factors of 6 are 1 and 6, and 2 and 3. Izydorczak 2014
4) I know the number parts of the binomials have to add up to 5, because in FOIL, the "OI" parts add up to the x-coefficient. Which of the factors of 6, that I noted in the step above, add up to 5? The answer to that is 2 and 3. 5) So the variable parts of the binomial factors are both x, and the number parts of the binomial factors are 2 and 3. Therefore my factors are (x + 2)(x + 3). 6) I always check my work by multiplying the factors I come up with to make sure I come up with the polynomial I started with. (x + 2)(x +3) = x2 + 5x + 6, so my factors are correct. This method can be used for any polynomial of the form ax2 + bx + c, but it can be tricky to use when a is a value other than 1. In that case, you have to do extra figuring to determine what combinations of the factors of a and the factors of c will add up to give b. In cases where a is not 1, it is recommended you use another factoring method, such as factoring by grouping, the box method, the diamond method, or the slide and divide method. Izydorczak 2014
x2+ 5x + 6 Look (#1) at sign! 1) The sign of the last term (6) is positive. Therefore the signs of the two binomial factors are the same (since multiplying like signs gives a positive number; if the sign had been negative, we know the binomial factors would have different signs). ( + ) ( + ) or ( - ) ( - ) Izydorczak 2014
x2+ 5x + 6 2) The variable parts of the binomial factors are both going to be x, because only x * x (the multiplication of the "firsts") will give me x2. ( + ) ( + ) X X Izydorczak 2014
x2+ 5x + 6 3) I know the number parts of the binomials have to be multiply to be 6, since I will be multiplying them together to get the last term of the polynomial. Factors of 6 are 1 and 6, and 2 and 3. ( + ) ( + ) Izydorczak 2014
x2+ 5x + 6 4) I know the number parts of the binomials have to add up to 5, because in FOIL, the "OI" parts add up to the x-coefficient. Which of the factors of 6, that I noted in the step above, add up to 5? The answer to that is 2 and 3. ( X+ ) (X+ ) 3 2 Izydorczak 2014
x2+ 5x + 6 5) So the variable parts of the binomial factors are both x, and the number parts of the binomial factors are 2 and 3. Therefore my factors are (x + 2)(x + 3). (x+2) (x+3) Izydorczak 2014
x2+ 5x + 6 6) I always check my work by multiplying the factors I come up with to make sure I come up with the polynomial I started with. (x + 2)(x +3) = x2 + 5x + 6, so my factors are correct. (x+2) (x+3) CK. x2 + 2x + 3x + 6 x2 +5x + 6 Izydorczak 2014
Tryone! Izydorczak 2014
x2 - 4x + 4 - - 2 ( ) ( ) X X 2 CK. x2 - 2x - 2x + 4 x2 -4x + 4 Izydorczak 2014