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
@Jhannybean @PaxPolaris
Which of these are factorable?
@dtan5457 idk
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?
x2 – 6x + \(\color{red}{9}\) x2 – 2x + \(\color{red}{4}\) x2 + 5x + 10 x2 + 4x + \(\color{red}{16}\)
From your answer choices, first analyze the constants.
You see that \(9 = 3^2~,~ 4 = 2^2 ~,~ 16 = 4^2\)
\(x^2 \implies (x)^2\)
iDont Get This 1 @Jhannybean
I think the first two as @Jhannybean wrote
Then you use your constant terms to find your middle term
maybe can be x^2+5x+10, because when I substitute x=1, I get 16 which is a perfect square!
?
@Jhannybean Please continue your line of explanation. Don't get interrupted.
@AlexandervonHumboldt2 @Abhisar
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.
So let's try option D.
D. \(x^2 +4x+16\) We found that \(16 = 4^2\). that means we have \(4~,~4\) as our factors.
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.
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.
not sure
Well, expand these functions.
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.
idk im trying
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 :)
|dw:1419718482638:dw|
Join our real-time social learning platform and learn together with your friends!