Lennie Davis

2021-12-31

Find the LCM of the given polynomial.
$4{x}^{3}-4{x}^{2}+x,2{x}^{3}-{x}^{2},{x}^{3}$

twineg4

Step 1: Given
$4{x}^{3}-4{x}^{2}+x,2{x}^{3}-{x}^{2},{x}^{3}$
Step 2: Solution
Factoring them
$4{x}^{3}-4{x}^{2}+x=x{\left(2x-1\right)}^{2}2{x}^{3}-{x}^{2}={x}^{2}\left(2x-1\right)$
${x}^{3}$
LCM is the product of all the factors which are common, and which are not common
So LCM is
${x}^{3}{\left(2x-1\right)}^{2}$
${x}^{3}{\left(2x-1\right)}^{2}$

Mollie Nash

for the first one,
it's $x\left(4{x}^{2}-4x+1\right)⇒x\left(2x-1\right)\left(2x-1\right)$
for the second one,
${x}^{2}\left(2x-1\right)$
for the third one,
${x}^{3}$
hence lcm for these three is just x

Vasquez

You have to factor each polynomial first;
$\begin{array}{}4{x}^{3}-4{x}^{2}+x=x\left(2x-1{\right)}^{2}\\ 2{x}^{3}-{x}^{2}={x}^{2}\left(2x-1\right)\\ {x}^{3}={x}^{3}\end{array}$
Then we see we need to have a total of two (2x-1) factors, since the highest power is 2, and three x factors, since the highest power is 3. The Least Common Multiple is then ${x}^{3}\left(2x-1{\right)}^{2}=4{x}^{5}-4{x}^{4}+{x}^{3}.$

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