factorise 3a^3+17a+10
Answers
Answer:3a2+17a+10=0
Two solutions were found :
a = -5
a = -2/3 = -0.667
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "a2" was replaced by "a^2".
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(3a2 + 17a) + 10 = 0
Step 2 :
Trying to factor by splitting the middle term
2.1 Factoring 3a2+17a+10
The first term is, 3a2 its coefficient is 3 .
The middle term is, +17a its coefficient is 17 .
The last term, "the constant", is +10
Step-1 : Multiply the coefficient of the first term by the constant 3 • 10 = 30
Step-2 : Find two factors of 30 whose sum equals the coefficient of the middle term, which is 17 .
-30 + -1 = -31
-15 + -2 = -17
-10 + -3 = -13
-6 + -5 = -11
-5 + -6 = -11
-3 + -10 = -13
-2 + -15 = -17
-1 + -30 = -31
1 + 30 = 31
2 + 15 = 17 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 2 and 15
3a2 + 2a + 15a + 10
Step-4 : Add up the first 2 terms, pulling out like factors :
a • (3a+2)
Add up the last 2 terms, pulling out common factors :
5 • (3a+2)
Step-5 : Add up the four terms of step 4 :
(a+5) • (3a+2)
Which is the desired factorization
Equation at the end of step 2 :
(3a + 2) • (a + 5) = 0
3a+2=0
3a=-2
a=-2/3
a+5=0
a=-5
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Step-by-step explanation: