Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

A perfect square trinomial can be represented by a square model with equivalent length and width. Which polynomial can be represented by a perfect square model? x2 – 6x + 9 x2 – 2x + 4 x2 + 5x + 10 x2 + 4x + 16

OpenStudy (anonymous):

@Jhannybean @PaxPolaris

OpenStudy (dtan5457):

Which of these are factorable?

OpenStudy (anonymous):

@dtan5457 idk

OpenStudy (dtan5457):

ax^2+bx+c=0 which of these has values where the same number can add up to value of b, and multiply to the value of c?

OpenStudy (jhannybean):

x2 – 6x + \(\color{red}{9}\) x2 – 2x + \(\color{red}{4}\) x2 + 5x + 10 x2 + 4x + \(\color{red}{16}\)

OpenStudy (jhannybean):

From your answer choices, first analyze the constants.

OpenStudy (jhannybean):

You see that \(9 = 3^2~,~ 4 = 2^2 ~,~ 16 = 4^2\)

OpenStudy (jhannybean):

\(x^2 \implies (x)^2\)

OpenStudy (anonymous):

iDont Get This 1 @Jhannybean

OpenStudy (michele_laino):

I think the first two as @Jhannybean wrote

OpenStudy (jhannybean):

Then you use your constant terms to find your middle term

OpenStudy (michele_laino):

maybe can be x^2+5x+10, because when I substitute x=1, I get 16 which is a perfect square!

OpenStudy (anonymous):

?

OpenStudy (mathmate):

@Jhannybean Please continue your line of explanation. Don't get interrupted.

OpenStudy (anonymous):

@AlexandervonHumboldt2 @Abhisar

OpenStudy (jhannybean):

Ok, then using your constant terms, you have to test whether one is positive, the other is negative, they are both positive, or both are negtive.

OpenStudy (jhannybean):

So let's try option D.

OpenStudy (jhannybean):

D. \(x^2 +4x+16\) We found that \(16 = 4^2\). that means we have \(4~,~4\) as our factors.

OpenStudy (jhannybean):

Now if we write it in factored form, does \((x+4)(x+4) = x^2 +4x+16\)? If it does not, then we can eliminate that option.

OpenStudy (jhannybean):

Next we analyze option B. \(x^2 -2x+4\). Recall that \(4 = 2^2\), and that leaves us with the factors \( 2~,~2\). writing this in factored form, we have \((x+2)(x+2)\) and even \((x-2)(x-2)\) Do \((x+2)(x+2) = x^2 -2x+4\) or does \((x-2)(x-2) = x^2 -2x+4\)? If not, then eliminate the choice.

OpenStudy (anonymous):

not sure

OpenStudy (jhannybean):

Well, expand these functions.

OpenStudy (jhannybean):

What do these functions expand to?: \[(x-3)(x-3)=\]\[(x+4)(x+4)=\]\[(x+2)(x+2)=\]\[(x-2)(x-2)=\]Use the foil method to expand all of these functions and tell me what you get.

OpenStudy (anonymous):

idk im trying

OpenStudy (jhannybean):

I'll help you with the last one, then I'd like you to use the same technique on the rest of them and expand them accordingly :)

OpenStudy (jhannybean):

|dw:1419718482638:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!