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Chapter 0: Problem 61

Find each product. $$(x-3 y)(2 x+7 y)$$

Short Answer

Expert verified
The product of \((x-3y)\) and \((2x+7y)\) is \(2x^2 + xy - 21y^2\).

Step by step solution

01

Multiply the First Two Terms

Firstly, multiply the first term of the first binomial by the first term of the second binomial. That is, multiply \(x\) and \(2x\) to get \(2x^2\).
02

Multiply the Outer Terms

Secondly, multiply the outer terms, which are the first term of the first binomial and the second term of the second binomial. That is, multiply \(x\) and \(7y\) to get \(7xy\).
03

Multiply the Inner Terms

Thirdly, multiply the inner terms, which are the second term of the first binomial and the first term of the second binomial. That is, multiply \(-3y\) and \(2x\) to get \(-6yx\) or \(-6xy\) (as the order of the variables does not matter in the product).
04

Multiply the Last Two Terms

Lastly, multiply the last term of the first binomial by the last term of the second binomial. That is, multiply \(-3y\) and \(7y\) to get \(-21y^2\).
05

Combine Like Terms

Finally, combine all those calculated values. The expression becomes \(2x^2 + 7xy - 6xy - 21y^2\). Now, you can combine the like terms \(7xy\) and \(-6xy\) to get \(xy\). So the final simplified expression will be \(2x^2 + xy - 21y^2\).

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