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.

04:00

### 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 squared plus or minus two ๐๐ plus ๐ squared, which can be factored as ๐ plus or minus ๐ squared.

In these trinomials, ๐ and ๐ may be variables, constants, or products of variables and constants. In this example, if we take ๐ 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 is the square root of ๐ 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๐, or in some cases negative two ๐๐. Two ๐๐ 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๐ 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 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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