What is (d^2+4d+3)(d+1)^{-1} ?

The solution to (d^2+4d+3)(d+1)^{-1} is d+3

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

simplify

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

d + 3

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

(d + 1)^{- 1} = \frac{1}{d + 1}

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

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Apply rule:

a \cdot 1 = a

(d^{2} + 4 d + 3) \cdot 1 = (d^{2} + 4 d + 3)

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

Factor

d^{2} + 4 d + 3 : (d + 1) (d + 3)

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

Cancel the common factor:

d + 1

= d + 3

(x + 1)^{3}

e^{2 x} \times e^{- 2 x}

500 \cdot e^{- 0.25 \cdot 10}

- 3 a^{6} b^{2} (2 a^{5} b^{7})