Video: US-SAT04S3-Q05-279193086407

Which of the following is equivalent to 4π‘Žβ΄ + 20π‘ŽΒ²π‘Β² + 25𝑏⁴? [A] (4π‘ŽΒ² + 25𝑏²)Β² [B] (4π‘Ž + 25𝑏)⁴ [C] (2π‘ŽΒ² + 5𝑏²)Β² [D] (2π‘Ž + 5𝑏)⁴

04:42

Video Transcript

Which of the following is equivalent to four π‘Ž to the fourth power plus 20π‘Ž squared 𝑏 squared plus 25𝑏 to the fourth Power? A) Four π‘Ž squared plus 25𝑏 squared squared, B) four π‘Ž plus 25𝑏 to the fourth power, C) two π‘Ž squared plus five 𝑏 squared squared, or D) two π‘Ž plus five 𝑏 to the fourth power.

If we look closely at this expression, we could rewrite four π‘Ž to the fourth power as two π‘Ž squared squared. And we could rewrite 25𝑏 to the fourth power as five 𝑏 squared squared. But what about the middle? I know that 20 equals two times two times five. And so I’m starting to see a pattern. We now have something that says two π‘Ž squared squared plus two times to π‘Ž squared times five 𝑏 squared plus five 𝑏 squared squared. This is good because we recognize the pattern. π‘Ž squared plus two π‘Žπ‘ plus 𝑏 squared is equal to π‘Ž plus 𝑏 squared. In our case, the capital 𝐴 would be equal to two π‘Ž squared. And the capital 𝐡 would be equal to five 𝑏 squared. And so we could say that our 𝐴 plus 𝐡 squared would be two π‘Ž squared plus five 𝑏 squared squared, which is option C.

Solving using this method relies upon you recognizing patterns and the expression we were given. But what if you didn’t recognize that or at least didn’t initially recognize that? Let’s take this expression and then consider the four answer choices using a square. Option A says four a squared plus 25𝑏 squared squared. And so we put four π‘Ž squared and 25𝑏 squared on the left. And on the top of our square, in the top left hand corner, we’ll multiply four π‘Ž squared times four π‘Ž squared, which gives us 16π‘Ž to the fourth. This is more than our original expression. And so we know that A will not work. In option B, we have four π‘Ž plus 25𝑏. But we’re taking it to the fourth power. This will be equal to four π‘Ž plus 25𝑏 times itself four times. This would result in a far greater value than the one we’ve been given.

Let’s consider option C one more time. It’s two π‘Ž squared plus five 𝑏 squared times two π‘Ž squared plus five 𝑏 squared. And we get four π‘Ž to the fourth power, 10 π‘Ž squared 𝑏 squared, two times five is 10 and π‘Ž squared times 𝑏 squared is just π‘Ž squared 𝑏 squared. We get the same thing in the bottom left, 10 π‘Ž squared 𝑏 squared. And then we have 25𝑏 to the fourth power. We recognize our four π‘Ž to the fourth power and our 25𝑏 to the fourth power. 10π‘Ž squared 𝑏 squared and 10π‘Ž squared 𝑏 squared are like terms and can be combined together to equal 20π‘Ž squared 𝑏 squared. And this confirms that C), two π‘Ž squared plus five 𝑏 squared squared, is equal to four π‘Ž to the fourth power plus 20π‘Ž squared 𝑏 squared plus 25𝑏 to the fourth power.

We would have two π‘Ž plus five 𝑏 times itself four times. Using the square method, you would multiply two π‘Ž plus five 𝑏 times itself. And then you would multiply whatever you found by itself again. And you would end up with too many terms for it to be equivalent to the expression we started with. And so our final answer is two π‘Ž squared plus five 𝑏 squared squared.

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