FACTORING HELP SHEET

 

There are many types of factoring problems.  This sheet is to help you with factoring but is not the only method for studying for the exam.  These are tips that I use when I factor.

Problems:

1.  6a4b2 + 9a3b  This problem is a binomial so look at the two terms and see if there is a common factor in the two coefficients.  If so, write that factor down.  Now look at the variables and see if any of them repeat.  If so, take the smallest exponent of that variable out.  In this problem we would take out 3 a3b.  Now we divide this factor into both factors and write the result in parentheses or we can look at it as what times the common factor to get the original problem.  Our answer is 3 a3b(2ab + 3).  Always check your final answer to be sure it can’t be factored any more.

 

2.  Problem 1 process can be applied to x(a + 3) – 4(a + 3) to get (a + 3)(x – 4).

 

3.  To factor trinomials, follow the following steps.  Ex.  x4 –  7x3 + 10x2

 

a.       As in problem 1, factor out any like terms.  x2 (x2 –7x + 10)

b.      Notice the leading coefficient is 1.  Put two sets of parentheses.  x2(     )(     )

c.       Put the variable inside the parentheses in the place where it is in the trinomial, (either first or last)  x2 (x     )(x    ).

d.      Look at the constant or last term sign.  If positive, the signs are the same since the product of two positives is positive and the product of two negatives is positive also.  Take the sign of the second term and put that in the parentheses.   If the third sign is negative, the signs will be opposite.

x2 (x -  ) (x  -  ).

e.       Find factors the last coefficient or term to apply the third term sign to get the middle term.  In this case 5 time 2 is 10 and 5 + 2 is 7 so place the factors in the appropriate place.  In this problem it doesn’t make any difference since both signs are negative.  x2 (x – 2)(x – 5).   Do your check to make sure the O and I of FOIL will give you the middle term.  The F and L should work if you factored correctly.

 

4.  To factor a trinomial with the leading coefficient is not one, 2x2 – x – 6. 

           

a.       This problem has no common factors so we do not have to do this step.

b.      Multiple the coefficients of the first and last term.  2 times 6 is 12.

c.       Take factors of 12 and apply the sign of the third term.  3 times 4 is 12 and

3 – 4 is –1, which is the coefficient of the second term.

d.      Rewrite the original problem but replacing the factors with the middle term, like 2x2 + 3x – 4x – 6. 

e.       Now we have a 4-term polynomial so we can factor the first two terms and the last two terms to get a problem similar to problem 2.  x(2x + 3) – 2(2x + 3).  After factoring you should come up with the two parentheses to be the same.

f.       Complete the factoring as in problem 2.  (2x + 3)(x – 2).

 

5.  Problem 5 is a polynomial with four terms like in 4d. Solve by using the step

 4e and 4f.

 

6. If you have a trinomial, similar to problem 4, check to see, after taking out all

common factors, if the first and last terms are both perfect squares and the sign of the third term is positive.  Example such as 25x2 + 30x + 9

 

a.   Make two sets of parentheses (    ) (    ).

  1. Put the square root of the first term at the beginning of each parentheses. 

(5x   )(5x   ).

c.   Put the sign of the second term in, (5x +   )(5x +  ).

d.   Put the square root of the last term in the remaining place.  (5x +3)(5x + 3)

e.   Check to see if the middle term works as described in 3e.

 

7.  If you have a binomial where each term is a perfect square after like terms are taken

 out.

 

  1. If there is a plus sign between the two terms, such as x2 + 4, this is not factorable.
  2. If there is a minus sign between the two terms, such as x2 – 4, then put two parentheses with opposite signs and fill in with the square roots of each term.

(x – 2)(x + 2)

 

8.  If you have a binomial with two perfect cubes, such as x3 – 8, follow the following

 steps.

 

      a.    Make two sets of parentheses, the first normal size the second larger in width.

  1.  Put the cube root of each term in the parentheses separated by the sign in the

 original problem.  (x – 2)(             )

  1. Square the first term of the new binomial and place in the first part of the second

      parentheses.  OR what times the first term of the new to get the first term of the

original.  This will be the same for both.  (x – 2)(x2      )

  1. Take the opposite of the sign in the original problem and put that after the first

      term in the second parentheses.  (x – 2)(x2 +     )

  1. Multiply the two terms, ignoring the sign between them, in the new binomial,

x times 2 and put this in for the second term in the second set of parentheses,

(x – 2)(x2 + 2x     ).

      f.    Put a plus sign next.  THIS WILL ALWAYS BE POSITIVE.  (x – 2)(x2 + 2x +   )

g.    Find what times the second term in the new binomial to get the second term in

 the original binomial.  It will be the square of the second term of the new

 binomial.  Put this in the last term of the second parentheses.  (x – 2)(x2 + 2x + 4)

 

YOU NEED TO KNOW THIS FOR THIS EXAM AND THE REST OF THIS COURSE AS WELL AS FOR COLLEGE ALGEBRA.

 

HAPPY FACTORING!