How to solve this? (adding polynomial fractions)I'm having trouble solving this expression:(x−1)(7x+6)(x−1)(x+1)2−7(x+1)What's the steps to solve this?I know you expand (x+1)2 to (x+1)(x+1)and that you need to find a common denominator before adding the numerators together.The final answer is −1x2+2x+1.Thanks.

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How to solve this? (adding polynomial fractions)I&#039;m having trouble solving this expression:(x−1)(7x+6)(x−1)(x+1)2−7(x+1)What&#039;s the steps to solve this?I know you expand  (x+1)2 to (x+1)(x+1)and that you need to find a common denominator before adding the numerators together.The final answer is −1x2+2x+1.Thanks.

anghoelv1lw · Accepted Answer

We want to simplify the following:(x−1)(7x+6)(x−1)(x+1)2−7(x+1)First, we cancel the common factor (x−1) in the left-hand fraction:(x−1)(7x+6)(x−1)(x+1)2−7(x+1)=(7x+6)(x+1)2−7(x+1)Now we find a common denominator to add the fractions, and see that we want both denominators to be (x+1)2. To accomplish this, we can multiply the numerator and the denominator of the second fraction by the factor of (x+1):(7x+6)(x+1)2−7(x+1)⋅(x+1)(x+1)=(7x+6)(x+1)2−7(x+1)(x+1)2=(7x+6−7(x+1))(x+1)2=−1(x+1)2=−1x2+2x+1

Riya Erickson · Accepted Answer

If x≠1  you can cancel (x−1) in the numerator and denominator of (x−1)(7x+6)(x−1)(x+1)2 so,(x−1)(7x+6)(x−1)(x+1)2−7(x+1)=7x+6(x+1)2−7(x+1)Now, if x≠−1 you can multiply and divide 7x+1 by (x+1) and add the fractions:7x+6(x+1)2−7(x+1)=7x+6(x+1)2−7(x+1)(x+1)2=7x+6−7(x+1)(x+1)2=−1(x+1)2

How to solve this? (adding polynomial fractions) I'm having

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