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chadandtricia_2009
What is the factored form of the expression? k2 – 9h2 A. (k + 3h)(k + 3h) B. h2(k + 3)(k – 3) C. (k + 3h)(k – 3h) D. (k – 3h2)(k + 3)
can you show me how you got that please?
i don't understand how to do it
K*K=K2 +3*-3=-9 H*H=H2
It is C. You have to first recognize the fact that this is a difference in squares pattern. \[a^{2} - b^{2} = (a + b)(a - b)\]
Which becomes k2-9h2
If you want to factor it without the pattern, then find factors of -9 that add up to 0. These factors are -3 and 3. THerefore, (k + 3h)(k – 3h)
In this problem you are dealing with perfect squares, so it was easier just to square root them. However, that only works if all of them are actually perfect squares. If you don't know what a perfect square is, it is a number multiplied by itself.
so what do I plug into the pattern?
a = k b = 3h See how k is squared and 3h is squared as well? k^2 = k^2 3^2h^2 = 9h^2
\[K timesK =K ^{2} 3h\times3h=9h /sqrt{9}=3\]
sorry, it didnt space out like I planned..
ok got it now.
thanks alot all of you
@zeesbrat3 and @doubledrive In the future, please don't just give the answer (sometimes people just want to cheat and that is not what OpenStudy is about.) Please guide the user so that the person can come to the answer themselves. Also @chadandtricia_2009 thanks for asking for an explanation. :)
And that message goes to anyone I miss that didn't know the new CoC. thanks.
@myininaya for your information, i answered the question and i explained it.
A. (k + 3h)(k + 3h) I got this correct on my test.