5v^3+4v^2-15v-12=(v^2 - 3) • (5v + 4) is the roots of cube equation
A divisor of an integer n, also called a factor of n, is an integer m that may be multiplied by some integer to produce n.
given a cube equation 5v^3+4v^2-15v-12
Equation at the end of step
((5v^3 + 4v^2) - 15v) - 12
Checking for a perfect cube
3.1 5v^3+4v^2-15v-12 is not a perfect cube
Trying to factor by pulling out :
3.2 Factoring: 5v^3+4v^2-15v-12
Thoughtfully split the expression at hand into groups, each group having two terms ,
Group 1,-15v-12
Group 2,5v^3+4v^2
Group 1, (5v+4) • (-3)
Group 2, (5v+4) • (v^2)
Add up the two group,(5v+4) • (v^2-3)
Which is the desired factorization
Trying to factor as a Difference of Squares,v^2-3
A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
(A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
AB = BA is the commutative property of multiplication.
- AB + AB equals zero and is therefore eliminated from the expression.
3 is not a square
Binomial can not be factored as the difference of two perfect squares. (v^2 - 3) • (5v + 4)
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