Okay, so I am supposed to simplify each radical and then combine. original equation: -3x*cbrt(54x) + 2*cbrt(16x^7). I factor the primes out of the radicands and rewrite the equations so no exponents are > the index. I then take out opposite operations(I would show work however I am not fluent in LaTex and it is too much for me to type.) After I do all of the necessary simplifying I end up with -9x*cbrt(2x) + 4x^2*cbrt(2x). Is this as far as I can go? I understand they are like radicals because the radicand and index are the same however, I am confused as to how I would combine the (-9x)+(4x^2). Or is what I ended up with the simplified answer? Thanks in advance.
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combining radical expressions!
- Thread starter beranabus
- Start date
\(\displaystyle -3x\sqrt[3]{54x} + 2\sqrt[3]{16x^7} = -3x\sqrt[3]{2 * 3^3 * x} + 2\sqrt[3]{2 * 2^3 * x^6 * x} = (-3x)(3)\sqrt[3]{2x} + 2 * 2x^2\sqrt[3]{2x} = - 9x\sqrt[3]{2x} + 4x^2\sqrt[3]{2x}.\) So far so good.Okay, so I am supposed to simplify each radical and then combine. original equation: -3x*cbrt(54x) + 2*cbrt(16x^7). I factor the primes out of the radicands and rewrite the equations so no exponents are > the index. I then take out opposite operations(I would show work however I am not fluent in LaTex and it is too much for me to type.) After I do all of the necessary simplifying I end up with -9x*cbrt(2x) + 4x^2*cbrt(2x). Is this as far as I can go? I understand they are like radicals because the radicand and index are the same however, I am confused as to how I would combine the (-9x)+(4x^2). Or is what I ended up with the simplified answer? Thanks in advance.
But you have common factors in your summands so you can simplify further. As you saw, the radicals are common because radicand and index are the same so
\(\displaystyle - 9x\sqrt[3]{2x} + 4x^2\sqrt[3]{2x} = \sqrt[3]{2x}(-9x + 4x^2) = \sqrt[3]{2x}(4x^2 - 9x).\)
But x is a common factor in x^2 and x.
\(\displaystyle \sqrt[3]{2x}(4x^2 - 9x) = x\sqrt[3]{2x}(4x - 9).\). Now you are done.
I gave you the answer because you are letting yourself be distracted by the radical. You would not have had trouble factoring
\(\displaystyle y(4x^2 - 9x)\ to\ xy(4x - 9).\) The radical just stands for a number; it does not change anything basic.