## Algebra Homework Help Step by Step

If we are given a problem we can analyze that and solve it using the required steps, for example if I consider a problem such as x

^{3 }- 27 and asked to find the roots of this equation. Then the step by step process can be evolved.

**Step 1:** x

^{3 }- 27

**Step 2:** x

^{3 }- 3

^{3}**Step 3:** Use a

^{3 }- b

^{3 }= (a - b) ( a

^{2 }+ ab + b

^{2})

**Step 4:** (x - 3)(x

^{2 }+ 3x + 9)

**Step 5:** x = 3

**Step 6:** using quadratic equation formula we can find the values of x

^{2 }+ 3x + 9

**Step 7:** -3 $\pm$ $\sqrt{\frac{9-36}{4}}$

**Step 8:** -3 $\pm$ 3 $\sqrt{\frac{3i}{4}}$

**Step 9:** The roots are X= 3 and two complex roots 3$\pm$3$\sqrt{\frac{3i}{4}}$

## Free Algebra Homework Help

Let us consider two examples in this section we will try to solve how to product two given matrices

Find the product AB if it exist for A = $\begin{bmatrix}

-1 & 3\\

4& -5\\

0& 2

\end{bmatrix}$ and B = $\begin{bmatrix}

1 & 2\\

0& 7

\end{bmatrix}$

A is a 3 x 2 matrix and B a 2 x 2 matrices.

Number of columns in matrix A = number of columns in matrix B = 2

Hence A and B are compatible for multiplication and the product AB is a matrix of size 3 x 2.

If AB = $\begin{bmatrix}

c_{11} & c_{12}\\

c_{21}& c_{22}\\

c_{31}& c_{32}

\end{bmatrix}$ then

$c_{11}= (-1)1 +3(0) =-1$ First row in A and First column in B

$c_{12}= (-1)2 +3(7) =19$ First row in A and second column in B

$c_{21}= 4(1) -5(0)=4$ Second row in A and first column in B

$c_{22}= 4(2)-5(7)=-27$ Second row in A and second column in B

$c_{31}= 0(1)+2(0)=0$ Third row in A and first column in B

$c_{32}= 0(2)+2(7)=14$ Third row in A and second column in B

AB = $\begin{bmatrix}

-1 & 19\\

4 & -27\\

0& 14

\end{bmatrix}$