Question

4x2+12x−3=8x344x−17x−3 4x2+2x−12x−3=(2x2+2x)+(−13x−3) = (x2+2x)+(−32x−3)=20(4t+1)+(−12t+x+1 2x(4x+1)+(−3x(4x+1)=(xx+2x(2x−1) 8x2−12t−1=8r2+2c−12x−1 2r2−11x−3=1x2−12r+2x−1. 82−17x+2x−1= a. Sin2=n=15 b 21r2−13t+2 =(C4−18−5C2y2+34) 4r3+12r3+1r−4=(4r4−12r2)+21r−4)=4r2(a−3)+3(r−2) a. 3k4−3k2−4+13 Factoring trinomials into binomials is your next step. If the trinomial has a leading coefficient of 1 , then we can easily use guess and check [or whatever method you prefer]. For example: Given the expression a2−6a+8, we can look at the factors of 8 that add up to −6 to determine how to factor the trinomial. Factors of 8 are 1&8 or 2&4, and because we need them to add up to −6, the factors muat both be negative. We can see that (−2)+(−4)=−6, so we have our factors, so a2−6a+8=(a−2)(a−4) 10. Factor the following expressions completely. Bex your final answers. a. n2+5n+6 b. x2−x−12 c. z2−2z−15 If the trinomial doesn't have a leading coefficient of 1 , then we can still use guess and check, but it usaally takes longer to determine the correet pairing. Using the A−C Method is a good alicnative. The A−C Method gets its name from the quadratic form a2+bx+c and the fact that we multiply a ×c to determine our factors. For example: Given the expression 8x2−10x−3, we first multiply 8×(−3)=−24. Next we list the fuctors of −24, and determine which pair will add to −10. Note that one factor will be positive and the other acgative. −1×24−2×12−3×8−4×61×(−24)2×(−12)3×(−8)4×(−6) We see that 2+(−12)=−10. Next, we re-write the middle term using the two factors as shown: 8x2−10x−3=8x2+2x−12x−3 This allows us to factor by grouping, We will group the first two terms and the last two terms. Then, factor out what is common: 8x2+2x−12x−3=(8x2+2x)+(−12x−3) "note that the negaive sign must be grouped with the 12 . which is the reason fic the + between the two groups (8x2+2x)+(−12x−3)=2x(4x+1)+(−3)(4x+1) Finally, we see that 4x+1 is common to both terms on the right. Factor it out to gct: 2x(4x+1)+(−3)(4x+1)=(4x+1)(2x−3) Second, is an example where factoring by grouping does NOT work. 4x3−12x2+3x−6=(4x3−12x2)+(3x−6)=4x2(x−3)+3(x−2) We are not able to continue factaring the expeession, because there are no common factors. 13. Factor each expression comple4ely using grouping. If it is not factorable, state so. a. 3k3−9k2−4k+12 b. m3+2m2−3m+2 c. 2x2y+6xy−4y−5x2−15x+10 14. Factor each expression completely using any method. If it is not factorable, state so. a. y2−16y+63 d. 4x2−36 b. 25x2−10x+1 c. 14bcd2+28bcd−42bc c. 25z3+125z2−z−5 f. 18y2+25y−3

Answer 1

To factor trinomials, use the guess and check method for trinomials with a leading coefficient of 1 and the A-C method for trinomials without a leading coefficient of 1.

Step-by-step explanation:

To factor the given expressions, we can use different methods depending on whether the trinomial has a leading coefficient of 1 or not. For trinomials with a leading coefficient of 1, we can use the guess and check method, while for trinomials without a leading coefficient of 1, we can use the A-C method. I will provide step-by-step solutions for the given examples:

1. a. n2 + 5n + 6 = (n + 2)(n + 3)
2. b. x2 - x - 12 = (x - 4)(x + 3)
3. c. z2 - 2z - 15 = (z - 5)(z + 3)

Learn more about Factoring trinomials