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calculusxy
 one year ago
factoring quadratic expression...
calculusxy
 one year ago
factoring quadratic expression...

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calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge 294a^2  54 \]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 @hartnn @Jhannybean

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0\[\large 294a^2  0a  54\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0larget common factor of \(294\) and \(24\) is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0get rid of that stupid zero

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0don't you mean the gcf of 294 and 54 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so start with \[6(49a^29)\] then factor \[49a^29\] as the difference of two squares

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a^2b^2 = (ab)(a+b)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or rather,.... \(ca^2b^2 = (cab)(ca+b)\) :') \(c\) is the coefficient of one of the terms lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\heartsuit^2\spadesuit^2=(\heartsuit+\spadesuit)(\heartsuit \spadesuit)\][

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh you think you're fancy huh

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\xi^2 \omega^2=(\xi+\omega)(\xi\omega)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You are correct @calculusxy , just don't forget to multiply by the 6

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0can i make sure the answer to another question?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0\[\large 12x^2 + 12x + 3\] \[\huge Answer \] \[\large 3(2x+1)(2x+1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0looks good to me you @Jhannybean ?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0well i did the same question, but in a different way and got a different answer ( can u catch my mistake) 12x^2 + 12x + 3 12x^2 + 6x + 6x + 3 2x(2x+3) 3(2x + 3) (2x  3)(2x + 3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align}12x^2+12x+3 &\rightarrow 4x^2+4x+1 \\&\rightarrow x^2+4x+4 \\&\rightarrow (x+2)(x+2) \\&\rightarrow \left(x+\frac{2}{4}\right)\left(x+\frac{2}{4}\right) \\&\rightarrow \left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right) \\&\rightarrow \color{red}{(2x+1)(2x+1)}\end{align} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For this method i simplify the quadratic by putting it in standard form \(ax^2+bx+c\) typically where \(a=1\) find the factors divide the factors by the leading coefficient we used to simplify our equation If the simplification of the factors does not give you an integer value but a fractional one, multiply the variable with the denominator of the fraction. You'll have your factorization

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0if there is a GCF better to take it out dw:1444172023392:dw

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[12x^2+6x > 6x(2x+1)\] these are the mistakes :=)
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