In Exercises 1–22, factor each difference of two squares. Assume ...

Question

In Exercises 1–22, factor each difference of two squares. Assume that any variable exponents represent whole numbers.

9x⁴ - 25y⁶

Accepted Answer

Hello, everyone. The following expression is a special product called the difference of two squares. We're gonna factorize it. The expression is 49 P raised to the fourth power minus 64 Q raised to the sixth power. Our answer choices are a, the parentheses P squared minus Q cubed multiplied by the parentheses. Seven P minus eight Q multiplied by the parentheses. Seven P plus eight Q B, the parentheses eight P squared minus seven Q cubed multiplied by the parentheses. Eight P squared plus seven Q cubed C Q multiplied by the parentheses. Seven P minus eight Q multiplied by the parentheses seven P plus eight Q D, the parentheses seven P squared minus eight Q cubed multiplied by the parentheses seven P squared plus eight Q cubed. Again, our original expression was 49 P raised to the 4th power -64. Q raised the 6th power. We are told that this is the difference of two squares, which means we know both terms here are perfect squares. I'm going to rewrite my variables so that I have a power raised to a power. So the 49 is Gonna stay put. And I recall that when I raise a power to a power, I keep the base and I multiply the exponents. So undoing that multiplication, I'm gonna divide the exponent of 4 x 2. So I now have in parentheses P squared and now it's squared -64. And I'm gonna do the same thing to the Q. I'm gonna divide the exponent of six by two. So I get Q to the third squared. So after this rewrite, I have 49 multiplied by P squared squared minus 64 multiplied by Q cubed squared. Now I recall the difference of two squares has me opening two sets of parentheses next to each other. And then I take the square root of the first term to find the first term in each new binomial. So the square root of 49 P squared squared, the square root of 49 is 7. The square root of P squared squared is P squared. So 7 p squared is the first term in each of our new binomials. The second term in our new binomials comes from the square root of the second term in the difference of two squares. So we're gonna do the square root of 64 Q cubed squared. The square root is 64 is 8. The square root of Q cubed squared is Q cubed. So 8 q cubed is the second term in both of our binomials. Lastly recall that when you are doing the difference of two squares in between our terms. And our new binomials one is a minus sign. One has a plus sign. So my factorization reads parentheses seven P squared minus eight Q cubed multiplied by the parentheses seven P squared plus eight Q cubed. I quickly glanced to see if I could factor this any further, but I do not have another difference of two squares. I have no greatest common factor. So this is my solution and it is answer choice. D have a nice day.

In Exercises 1–22, factor each difference of two squares. Assume that any variable exponents represent whole numbers. 9x⁴ - 25y⁶

Key Concepts

Difference of Squares

Factoring Polynomials

Exponents and Variables

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