(10a^2 +53a -37) ÷ (10a-7)
Answers
Answer:
STEP
1
:
Equation at the end of step 1
(32a2 + 53a) - 70
STEP
2
:
Trying to factor by splitting the middle term
2.1 Factoring 9a2+53a-70
The first term is, 9a2 its coefficient is 9 .
The middle term is, +53a its coefficient is 53 .
The last term, "the constant", is -70
Step-1 : Multiply the coefficient of the first term by the constant 9 • -70 = -630
Step-2 : Find two factors of -630 whose sum equals the coefficient of the middle term, which is 53 .
-630 + 1 = -629
-315 + 2 = -313
-210 + 3 = -207
-126 + 5 = -121
-105 + 6 = -99
-90 + 7 = -83
-70 + 9 = -61
-63 + 10 = -53
-45 + 14 = -31
-42 + 15 = -27
-35 + 18 = -17
-30 + 21 = -9
-21 + 30 = 9
-18 + 35 = 17
-15 + 42 = 27
-14 + 45 = 31
-10 + 63 = 53 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -10 and 63
9a2 - 10a + 63a - 70
Step-4 : Add up the first 2 terms, pulling out like factors :
a • (9a-10)
Add up the last 2 terms, pulling out common factors :
7 • (9a-10)
Step-5 : Add up the four terms of step 4 :
(a+7) • (9a-10)
Which is the desired factorization
Final result :
(9a - 10) • (a + 7)
Answer:
10a^2 +53a -37) ÷ (10a-7)