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For what real values of $c$ is $x^2 - 8x - 4x + c + x^2 - 20x + x^2$ the square of a binomial?

 Jun 9, 2024

Best Answer 

 #1
avatar+1790 
+1

First, let's combine all like terms. 

 

We have \(3x^{2}-32x+c\) as the simplified version. 

 

Now, let's divide everything by 3 to make our lives idea. 

 

We get \(x^2-\frac{32}{3}x+c/3\)

 

Now, we divide 32/3 by 2, and we get \(16/3\)

 

Now, \((16/3)^2 = 256/9\)

 

Now, we have 

\(c/3 = 256/9\\ c=253/3\)

 

Thanks! :)

 Jun 9, 2024
 #1
avatar+1790 
+1
Best Answer

First, let's combine all like terms. 

 

We have \(3x^{2}-32x+c\) as the simplified version. 

 

Now, let's divide everything by 3 to make our lives idea. 

 

We get \(x^2-\frac{32}{3}x+c/3\)

 

Now, we divide 32/3 by 2, and we get \(16/3\)

 

Now, \((16/3)^2 = 256/9\)

 

Now, we have 

\(c/3 = 256/9\\ c=253/3\)

 

Thanks! :)

NotThatSmart Jun 9, 2024

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