How do you factor completely #18x^2-30x-3x+5#?

Question

Kennedy Adams · Accepted Answer

y = (6x - 1)(3x - 5)

I use the systematic new AC Method to factor trinomials #f(x) = 18x^2 - 33x + 5 = #18(x + p)(x + q) Converted trinomial: #f'(x) = x^2 - 33x + 90 =# (x + p')(x + q'). p' and q' have same sign since ac > 0. Factor pairs of (ac = 90) --> (2, 45)(3, 30). This sum is 33 = -b. Then the opposite sum (-3, -30) gives p' = -3 and q' = -30. Back to original trinomial: #p = (p')/a = -3/18 = -1/6# and #q = (q')/a = -30/18 = -5/3.# Factored form: #y = 18(x - 1/6)(x - 5/3) = (6x - 1)(3x - 5)#

Julia Adams · Answer

undefined To factor completely the expression $18x^2 - 30x - 3x + 5$, you first group the terms:

$= (18x^2 - 3x) + (-30x + 5)$

Then, factor out the greatest common factor from each group:

$= 3x(6x - 1) - 1(6x - 1)$

Now, you can see that both groups have a common factor of $6x - 1$, so you can factor it out:

$= (3x - 1)(6x - 1)$

Therefore, the expression $18x^2 - 30x - 3x + 5$ factors completely to $(3x - 1)(6x - 1)$.

Nathan Axel · Answer

undefined To factor completely $18x^2 - 30x - 3x + 5$, first, group the terms:

$18x^2 - 30x - 3x + 5 = (18x^2 - 30x) + (-3x + 5)$

Then, factor out the greatest common factor from each group:

$6x(3x - 5) - 1(3x - 5)$

Now, notice that both terms share a common factor of $3x - 5$. Factor it out:

$(3x - 5)(6x - 1)$

So, the completely factored form of $18x^2 - 30x - 3x + 5$ is $(3x - 5)(6x - 1)$.