$2\left(3{r}^{2}+4r+2\right)-3\left(-{r}^{2}+4r-5\right)$ the problem already answered was written out wrong so please helpme with this one.
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eishale2n
$2\left(3{r}^{2}+4r+2\right)-3\left(-{r}^{2}+4r-5\right)$I take it we are just going tosimplify.
We distribute the 2 to the left term and distribute the -3 to theright term. We are careful when distributing the -3 because thesigns will change.
$6{r}^{2}+8r+4+3{r}^{2}-12r+15$ (Now we are just going toadd/subtract like terms)
$9{r}^{2}-4r+19$ We cannot simply further becausethe trinomial won't factor.
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Ciara Rose
take the 2 and multiply for all the factors tahat are in the () and take the number 3 and multiply with all the factors inthe ( ), remember (-)multiply with other negative will be (+),therefore the solution is:
$2\left(3{r}^{2}+4r+2\right)-3\left(-{r}^{2}+4r-5\right)$ $6{r}^{2}+8r+4+3{r}^{2}-12r+15$, now search the common factors( samevariable)
$9{r}^{2}-4r+19$ and these is the final solution,because this expression can´t be simplified