Question Video: Factorizing Algebraic Expressions by Taking out the GCF | Nagwa Question Video: Factorizing Algebraic Expressions by Taking out the GCF | Nagwa

Question Video: Factorizing Algebraic Expressions by Taking out the GCF Mathematics • First Year of Preparatory School

By taking out the HCF, factor the expression 14π‘₯⁡𝑦² βˆ’ 4π‘₯³𝑦 + 8π‘₯²𝑦.

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Video Transcript

By taking out the HCF, factor the expression 14π‘₯ to the fifth power 𝑦 squared minus four π‘₯ cubed 𝑦 plus eight π‘₯ squared 𝑦.

We’re given the expression 14π‘₯ to the fifth power 𝑦 squared minus four π‘₯ cubed 𝑦 plus eight π‘₯ squared 𝑦, and we need the HCF, the highest common factor. We’ll start by considering the highest common factor of the coefficients of these three terms. 14 equals two times seven. Four equals two times two. And eight equals two times four. The common factor here is two. It is true that eight and four share a factor of four, but 14 is not divisible by four. And so, we say that the common factor for all three coefficients is two. And that means we’ll rewrite the expression as two times seven π‘₯ to the fifth power 𝑦 squared minus two π‘₯ cubed 𝑦 plus four π‘₯ squared 𝑦.

From here, we consider the common factor of π‘₯. The third term has the smallest factor of π‘₯, π‘₯ squared, which means π‘₯ squared is the biggest factor of π‘₯ we can remove. So, we’ll undistribute a factor of π‘₯ squared from these three terms. π‘₯ to the fifth power divided by π‘₯ squared equals π‘₯ cubed. And we leave the 𝑦 squared. π‘₯ cubed divided by π‘₯ squared equals π‘₯ to the first power. And four π‘₯ squared 𝑦 divided by π‘₯ squared will be equal to four 𝑦. We now have a second equivalent expression. However, we still do not have our highest common factor. And we know this because all three of our terms have at least one factor of 𝑦. The smallest factor of 𝑦 here is 𝑦 to the first power. And that means that’s the most we can remove from all three terms. We undistribute 𝑦 to the first power from all three terms.

Our first term becomes seven π‘₯ cubed 𝑦 to the first power. Our second term is then negative two π‘₯ to the first power. When we remove a factor of 𝑦 to the first power from our third term, we’re just left with four. What we see now is that there are no common factors in the parentheses. And that means the highest common factor is what we’ve taken out. By undistributing the highest common factor two π‘₯ squared 𝑦, we have a fully factorized expression, two π‘₯ squared 𝑦 times seven π‘₯ cubed 𝑦 minus two π‘₯ plus four. If you wanted to check if this was true, you would multiply the highest common factor back by the three remaining terms, which would give you the expression you started with.

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