Q15. - Algebra 1

Question

Find the product.

$\frac{{\mathbf{a}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{a}\mathbf{+}\mathbf{1}}{\mathbf{a}\mathbf{+}\mathbf{1}}\mathbf{.}\frac{\mathbf{a}\mathbf{-}\mathbf{1}}{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}$

Step-by-Step Solution

Answer

The final answer is 1.

Step 1. Define a rational function.

A function of the form $y=\frac{p}{q}$ is called a rational function, where $p$ and $q$ are the polynomials and $q\ne 0$.

Step 2. Define the product of two expressions a b and c d .

The product of two expressions $\frac{a}{b}$ and $\frac{c}{d}$, where $b\ne 0$ and $d\ne 0$, is given below.

$\frac{a}{b}.\frac{c}{d}=\frac{ac}{bd}\mathrm{...........}\left(1\right)$

Step 3. Calculate the product a 2 + 2 a + 1 a + 1 . a − 1 a 2 − 1 .

Use $\left(1\right)$ to find the product  $\frac{a^2+2a+1}{a+1}.\frac{a-1}{a^2-1}$.

$\begin{array}{l}\frac{a^2+2a+1}{a+1}.\frac{a-1}{a^2-1}=\frac{(a^2+2a+1)(a-1)}{(a+1)(a^2-1)}\\ =\frac{(a^2+2\left(a\right)\left(1\right)+1^2)(a-1)}{(a+1)(a^2-1^2)}\\ =\frac{{(a+1)}^2(a-1)}{(a+1)(a+1)(a-1)}\\ =\frac{{(a+1)}^2(a-1)}{{(a+1)}^2(a-1)}\\ =1\end{array}$

Therefore, the final answer is 1.