This week in Precalc 11 we learned more about solving quadratic equation with factoring. We used our previous knowledge from the unit before on factoring and used it to hep us with solving quadratic equations. To be able to tell whether an equation is linear or quadratic is with the degree or exponent. You can tell the it is a quadratic equation if the degree is 2, and it is linear if there is no exponent. So, for example, 6x=0 is linear and x^2+3x-28=0 is quadratic.

When factoring a quadratic equation you will always have two answers for x, so if you have an equation, x^2-12x+32=0, the first thing you would do is factor the left side which would give you (x-8)(x-4)=0. If you look at this question, you are most likely to solve by using inspection and noticing that x has to be +8 and +4 for it to equal 0, but if not then you can solve each binomial individually by doing x-8=0 and x-4=0 which would give you x=8 and x=4.

There also may be questions the you have to distribute with first and then solve:

Sometime you may have and equation that is unable to factor, for example x^2+10x+12=0, there are no two numbers that multiply to 12 and add to 10. If this happens you have to change the x^2+10x into a perfect square. To do that you divide the 10 by 2 and then multiply that by itself. So it would be 10/2=5, 5×5=25. The next step would be be to write out the new equation, x^2+10x+25-25+12=0. You have to add +25 and -25 to make it even. Now you would factor x^2+10x+25, which gives you (x+5)^2, and then you would do -25+12, which gives you -13. So now the equation is (x+5)^2-13=0. Then you move the 13 to the right side making it positive. Once you do that you square root both sides to get rid of the exponent. Then you would move the 5 to the other side to isolate x and your final answer will be x=-5+-square root of 13. Here is the question written out to show the steps: