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Monday, October 21, 2013

Special Products - Factoring, 10

Category: Algebra

"Published in Newark, California, USA"

Find the factors for 


Solution: 

Consider the given equation above


If you think that the above equation cannot be factored, then you must consider first the investigation of each terms whether they can be factored or not. The variables at the first and last terms are perfect square. Since the last term is negative, then obviously we cannot take a square root of a negative number and hence, the given equation is not a perfect trinomial square. We can check the above equation using discriminant if it can be factored or not as follows


where a, b, and c are the coefficients of a trinomial. Now, let's check the given equation as follows







Since the value of discriminant is a whole number, then the given equation can be factored. Next, we have to think the factors of the last term so that we add the two factors, it will be the same as the middle term. The possible factors of the last term are 1, -1, 135, -135, 5, -5, 27, -27, 3, -3, 45, -45, 9, -9, 15, and -15. If the middle term is -134, then the factors must be 1 and -135. When you add 1 and -135, it will give us -134. Therefore, the factors of the above equation are



Sunday, October 20, 2013

Special Products - Factoring, 9

Category: Algebra

"Published in Newark, California, USA"

Find the factors for


Solution:

Consider the given equation above


The first and the last terms are both perfect squares. The middle term is twice the product of the square roots of first and last terms. Let's check the middle term as follows


Since the middle term is exactly the same as the computation above, then the given trinomial is a perfect trinomial square. If the middle term is negative, then the square root of the last term must be negative. Therefore, the factors are

  






Therefore, the factors are

 

Saturday, October 19, 2013

Special Products - Factoring, 8

Category: Algebra

"Published in Newark, California, USA"

Find the factors completely for


Solution:

Consider the given equation above


Take out their common factor, we have








Since the two grouped terms are the difference of two squares, then we need to factor the two grouped terms as follows



Therefore, the factors are