Simplify (d^2+4d+3)(d+1)^-1

$(d^{2} + 4 d + 3) {(d + 1)}^{- 1}$

Step 1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$(d^{2} + 4 d + 3) \frac{1}{d + 1}$

Step 2

Multiply $d^{2} + 4 d + 3$ by $\frac{1}{d + 1}$ .

$\frac{d^{2} + 4 d + 3}{d + 1}$

Step 3

Factor $d^{2} + 4 d + 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

$\frac{(d + 1) (d + 3)}{d + 1}$

Step 4

Cancel the common factor of $d + 1$ .

Cancel the common factor.

$\frac{(d + 1) (d + 3)}{d + 1}$

Divide $d + 3$ by $1$ .

$d + 3$

Simplify (d^2+4d+3)(d+1)^-1

Related Questions

Support LCMGCF.com by disabling your adblocker.