<p> FACTORING TRINOMIALS – THE “GUESS-AND-CHECK” METHOD (FACTORING TRINOMIALS OF THE FORM ax 2  bx  c , and a  1 )</p><p>Given a quadratic expression with the terms ax 2  bx  c , we are often asked to factor. What we are being asked to do is find two binomials which multiply to give the original expression.</p><p>GENERAL STEPS EXAMPLE </p><p>STEP #1 : Write out the trinomial in descending Ex.) x 2  7x 12 order: ax 2  bx  c .</p><p>STEP #2 : Set up two empty sets of parentheses x 2  7x 12 below the polynomial. ( ) ( )</p><p>STEP #3 : The first numbers ( ) ( ) x 2  7x 12 x x  because x  x  x 2 must multiply together to equal the first term, ax 2 .</p><p>STEP #4 : The second numbers ( ) ( ) x 2  7x 12 x  3x  4 must multiply together to equal the last term, c, and also add together to equal the because (3)(4)  12 coefficient of the middle term, b. and (3)  (4)  7</p><p>2 STEP #5 : Check your answer using FOIL. x  3x  4  x  4x  3x 12 When you multiply using FOIL, you  x 2  7x 12 should get the original polynomial.</p><p>So, the conclusion is the factored form of x 2  7x 12 is x  3x  4 and x 2  7x 12 = x  3x  4. FACTORING TRINOMIALS – THE AC METHOD (FACTORING TRINOMIALS OF THE FORM ax 2  bx  c , and a  1 )</p><p>Given a quadratic expression with the terms ax 2  bx  c , we are often asked to factor. What we are being asked to do is find two binomials which multiply to give the original expression.</p><p>The AC Method is a method of factoring trinomials in the form ax 2  bx  c . This is an alternative to the “guessing method.”</p><p>GENERAL STEPS EXAMPLE </p><p>STEP #1 : Factor any common terms (if possible). Ex.) 6x 2 11x 10 Then identify a, b, and c. a  6 ; b  11; c  10</p><p>STEP #2 : Multiply a c . a c  6(10)  60</p><p>STEP #3 : Use the TI-83 Plus to set up the following on the Y= screen. a c Let Y  and let Y  x  Y 1 x 2 1 (In this step we are finding the factors of ac and the sum of the factors.)</p><p>STEP #4 : Use the TI-83 Plus table feature to find </p><p> the b value under the Y2 column.</p><p>(In this step we are finding the two factors Y2  11 of ac that sum to b.)</p><p>So, we will want to use 4 and –15 from the X</p><p> and Y1 columns.</p><p>STEP #5 : Rewrite the middle term (bx) using the 6x 2  4x 15x 10 values found in step #4 as coefficients. (Note that we have not changed the value of the polynomial.)</p><p>2 STEP #6 : Group the equation from step #5 into 6x  4x 15x 10 two separate parts. Then, factor by 2x3x  2  53x  2 grouping. 3x  22x  5</p><p>STEP #7 : Check your answer in step #6 by 3x  22x  5 multiplying the two factors using FOIL.  6x 2 15x  4x 10  6x 2 11x 10 which was the original problem. So, we now know that the factors of 6x 2 11x 10 are 3x  22x  5 and 6x 2 11x 10 = 3x  22x  5</p>