Question Video: Factorizing by Completing the Square | Nagwa Question Video: Factorizing by Completing the Square | Nagwa

Question Video: Factorizing by Completing the Square Mathematics • Second Year of Preparatory School

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Factor 16๐‘ฅโด๐‘ฆยฒ + 4๐‘ฆยฒ๐‘งโด by completing the square.

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Video Transcript

Factor 16๐‘ฅ to the fourth power ๐‘ฆ squared plus four ๐‘ฆ squared ๐‘ง to the fourth power by completing the square.

To begin, we will consider whether the two terms in the given polynomial have a highest common factor, or HCF, which may contain variables, constants, or products of variables and constants.

We determine that four ๐‘ฆ squared is the highest common factor of the two terms. By dividing each term by the HCF, we find the remaining terms in the parentheses to be four ๐‘ฅ to the fourth power plus ๐‘ง to the fourth power. We want to factor this expression by completing the square. So, we need to manipulate it to include a perfect square trinomial in the form ๐‘Ž squared plus or minus two ๐‘Ž๐‘ plus ๐‘ squared, which can be factored as ๐‘Ž plus or minus ๐‘ squared.

In these trinomials, ๐‘Ž and ๐‘ may be variables, constants, or products of variables and constants. In this example, if we take ๐‘Ž squared to be four ๐‘ฅ to the fourth power and ๐‘ squared to be ๐‘ง to the fourth power, then our value of ๐‘Ž is the square root of ๐‘Ž squared, which is equal to two ๐‘ฅ squared. And our value of ๐‘ is the square root of ๐‘ squared, which is equal to ๐‘ง squared. Then, our middle term is equal to two ๐‘Ž๐‘, or in some cases negative two ๐‘Ž๐‘. Two ๐‘Ž๐‘ comes out to two times two ๐‘ฅ squared times ๐‘ง squared, which is four ๐‘ฅ squared ๐‘ง squared.

In our next step, we will introduce the two ๐‘Ž๐‘ term into the original expression. For any term we introduce into the expression, we must add the same term with the opposite sign; this way, we are effectively adding zero, which does not change the polynomial. In this case, the zero gets added to the polynomial in the form of four ๐‘ฅ squared ๐‘ง squared minus four ๐‘ฅ squared ๐‘ง squared. Our expression with these new terms is four ๐‘ฆ squared times four ๐‘ฅ to the fourth power plus four ๐‘ฅ squared ๐‘ง squared plus ๐‘ง to the fourth power minus four ๐‘ฅ squared ๐‘ง squared. We can now factor the first three terms in the parentheses as a perfect square trinomial, giving us two ๐‘ฅ squared plus ๐‘ง squared squared.

Now we have a difference of squares, since the expression within the parentheses is being squared and four ๐‘ฅ squared ๐‘ง squared is a perfect square, specifically the square of two ๐‘ฅ๐‘ง, where ๐‘Ž is in the first parentheses and ๐‘ is in the second parentheses. Following the formula for factoring a difference of squares, we get two ๐‘ฅ squared plus ๐‘ง squared minus two ๐‘ฅ๐‘ง times two ๐‘ฅ squared plus ๐‘ง squared plus two ๐‘ฅ๐‘ง.

We then check whether the resulting polynomials within each set of parentheses can be factored. In this case, both polynomials are prime. Therefore, we have that four ๐‘ฆ squared times two ๐‘ฅ squared minus two ๐‘ฅ๐‘ง plus ๐‘ง squared times two ๐‘ฅ squared plus two ๐‘ฅ๐‘ง plus ๐‘ง squared represents the full factorization of 16๐‘ฅ to the fourth power ๐‘ฆ squared plus four ๐‘ฆ squared ๐‘ง to the fourth power.

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