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How do you divide #(x^2+6x+5) / (x+5)#?

Answer 1

Benjamin Achtenberg

#(x^2 + 6x + 5)/(x+ 5) = x+1#

Here, we can factor the numerator.

#(x^2 + 6x + 5)/(x+ 5) =((x+5)(x+1))/(x+5)#

Now we can cancel #x+5# from numerator and denominator.

#(x^2 + 6x + 5)/(x+ 5) =(cancel((x+5))(x+1))/(cancel(x+5))#

#(x^2 + 6x + 5)/(x+ 5) = x+1#

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Answer 2

Lincoln Anderson

To divide the polynomial ( (x^2 + 6x + 5) ) by ( (x + 5) ), you can use polynomial long division or synthetic division. Let's use polynomial long division:

Divide the first term of the dividend by the first term of the divisor: ( \frac{x^2}{x} = x ).
Multiply the entire divisor ( (x + 5) ) by the quotient term ( x ), and subtract the result from the dividend ( (x^2 + 6x + 5) ): ( (x^2 + 6x + 5) - (x^2 + 5x) = x ).
Bring down the next term from the dividend: ( +0x ).
Repeat the process: divide the first term of the new dividend (( x )) by the first term of the divisor (( x )), which results in ( 1 ).
Multiply the entire divisor ( (x + 5) ) by the new quotient term ( 1 ), and subtract the result from the new dividend: ( (x) - (x) = 0 ).

Therefore, the result of ( (x^2 + 6x + 5) ) divided by ( (x + 5) ) is ( x + 1 ), with a remainder of ( 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.