Did you notice that the terms can be grouped in 2 different ways yet result in

# Did you notice that the terms can be grouped in 2

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Did you notice that the terms can be grouped in 2 different ways yet result in the same two factors? Let’s try two more examples that involve the subtraction of some terms. Example 3 Factor by grouping: ax + bx ay by Step 1 ax + bx ay - by group terms in pairs Step 2 = x(a + b) y(a + b) extract GCF from each pair Step 3 = (a + b) (x y) extract binomial common factor ? Group terms together that have a common factor! NB: (a + b) = (b + a) NB When both terms have a negative constant, factor out a negative. This changes the sign of both terms. 10C P4-5 LG Factoring Page 13 of 35 Remember, there should be another grouping of terms that will also work! Step 1 ax ay + bx by Step 2 = a(x y) + b(x y) Step 3 = (x y) (a + b) Did you notice that we do not have to worry about signs with this grouping? If possible, when factoring by grouping, try to group terms so that the third sign is a plus sign ! One last example before you try some on your own. Example 4 Factor by grouping: 3x 2 + 9xy 2xy 6y 2 if possible, let’s keep the third sign a plus sign when grouping! Step 1 3x 2 2xy + 9xy 6y 2 group terms in pairs Step 2 = x(3x 2y) + 3y(3x 2y) extract GCF from each pair Step 3 = (3x 2y)(x + 3y) extract binomial common factor Check these last 2 examples on your own by multiplying the factors. Using your calculator to verify factoring To verify whether the factors are correct, follow the procedure: Step 1: Enter the original polynomial into “Y1 =”. Step 2: Enter the factored version into “Y2 =”. Step 3: Execute the Graph command and observe carefully. Step 4: If you see two distinct graphs, then the polynomial has been incorrectly factored. If the calculator graphs one graph on top of the other, then the polynomial has been graphed correctly. 10C P4-5 LG Factoring Page 14 of 35 Practice Exercise Factor the following polynomials using the factoring by grouping method. 1. 2mn + n 2 + 4m + 2n 6. bx + b + x + 1 2. ns + ms + nr + mr 7. xy y + 2x 2 3. mr + ms nr ns 8. 6x 2 3x + 2xy -y 4. 2m 2 + 4m 3m -6 *9. x 3 + x 1 5. abx 2 axy bxy + y 2 *10. 5y 3x + 3x² 5xy Answers to Practice Exercise 1. 2mn + n 2 + 4m + 2n = 2mn + n 2 + 4m + 2n = n(2m + n) + 2(2m+n) = (2m + n)(n + 2) 6. bx + b + x + 1 = bx + b + x + 1 = b(x + 1) + 1(x + 1) = (x + 1)(b + 1) 2. ns + ms + nr + mr = ns + ms + nr + mr = s(n + m) + r(n + m) = (n + m)(s + r) 7. xy - y + 2x - 2 = xy - y + 2x - 2 = y(x - 1) + 2(x - 1) = (x - 1)(y + 2) 3. mr + ms - nr - ns = mr + ms - nr - ns = m(r + s) - n(r + s) = (r + s)(m - n) 8. 6x 2 - 3x + 2xy - y = 6x 2 - 3x + 2xy - y =3x(2x - 1) + y(2x - 1) =(2x - 1)(3x + y) 4. 2m 2 + 4m - 3m - 6 = 2m 2 + 4m - 3m - 6 = 2m(m + 2) - 3(m + 2) = (m + 2)(2m - 3) 9. x 3 + x - x² - 1 = x 3 + x - x² - 1 = x(x 2 +1) - 1(x 2 + 1) = (x 2 + 1)(x - 1) 5. abx 2 - axy - bxy + y 2 = abx 2 - axy - bxy + y 2 = ax(bx - y) - y(bx - y) = (bx - y)(ax - y) 10. 5y - 3x + 3x² - 5xy = 3x 2 - 3x - 5xy + 5y (rearrange) = 3x 2 - 3x - 5xy + 5y = 3x(x - 1) - 5y(x - 1) = (x - 1)(3x - 5y) 