Question

trying to factor (4x-1)^1/2 - 1/3(4x-1)^3/2
Got nowhere! need full explanation on how to factor this. Tks

Answer 1

you have to factor out what is common, (4x-1)^1/2 is common to both terms

Answer 2

As LagrangeSon678 suggested, if you let:\[u=(4x-1)^{1/2}\]then you end up with:\[u-\frac{1}{3}u^3\]can you see how to factor this now?

Answer 3

have been 10 years out of school. Cannot remember the rules.

Answer 4

can you show the entire work step by step so i can understand? Thank you!

Answer 5

can you explain how did you get the U3 part (u-1/3 u3)?

Answer 6

can you show the entire work??

Answer 7

using the law of indicies, i.e.:\[a^{b/c}=(a^{1/c})^b\]we get:\[(4x-1)^{3/2}=((4x-1)^{1/2})^3\]so, if we substitute:\[u=(4x-1)^{1/2}\]we get:\[(4x-1)^{1/2}-\frac{1}{3}(4x-1)^{3/2}=(4x-1)^{1/2}-\frac{1}{3}((4x-1)^{1/2})^3=u-\frac{1}{3}u^3\]\[\qquad=\frac{u}{3}(3-u^2)\]then substitute the expression for u back into this to get:\[\frac{(4x-1)^{1/2}}{3}(3-((4x-1)^{1/2})^2)=\frac{(4x-1)^{1/2}}{3}(3-(4x-1))\]\[\qquad=\frac{(4x-1)^{1/2}}{3}(3-4x+1)=\frac{(4x-1)^{1/2}}{3}(4-4x)\]\[\qquad=\frac{4(4x-1)^{1/2}}{3}(1-x)=\frac{4(1-x)(4x-1)^{1/2}}{3}\]

Answer 8

Thank you! Now I got it.

Answer 9

yw - I'm glad I was able to help