calculusxy
  • calculusxy
factoring quadratic expression...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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calculusxy
  • calculusxy
\[\huge 294a^2 - 54 \]
calculusxy
  • calculusxy
@Nnesha
calculusxy
  • calculusxy
@jim_thompson5910 @hartnn @Jhannybean

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calculusxy
  • calculusxy
\[\large 294a^2 - 0a - 54\]
anonymous
  • anonymous
larget common factor of \(294\) and \(24\) is?
anonymous
  • anonymous
get rid of that stupid zero
calculusxy
  • calculusxy
okay sorry
calculusxy
  • calculusxy
don't you mean the gcf of 294 and 54 ?
anonymous
  • anonymous
yes i get 6
Jhannybean
  • Jhannybean
Hahahaha
anonymous
  • anonymous
so start with \[6(49a^2-9)\] then factor \[49a^2-9\] as the difference of two squares
Jhannybean
  • Jhannybean
\[a^2-b^2 = (a-b)(a+b)\]
calculusxy
  • calculusxy
(7a + 3)(7a - 3)
Jhannybean
  • Jhannybean
or rather,.... \(ca^2-b^2 = (ca-b)(ca+b)\) :') \(c\) is the coefficient of one of the terms lol
anonymous
  • anonymous
\[\heartsuit^2-\spadesuit^2=(\heartsuit+\spadesuit)(\heartsuit -\spadesuit)\][
Jhannybean
  • Jhannybean
Oh you think you're fancy huh
anonymous
  • anonymous
just showing off
Jhannybean
  • Jhannybean
lol
anonymous
  • anonymous
\[\xi^2 -\omega^2=(\xi+\omega)(\xi-\omega)\]
Jhannybean
  • Jhannybean
You are correct @calculusxy , just don't forget to multiply by the 6
calculusxy
  • calculusxy
can i make sure the answer to another question?
calculusxy
  • calculusxy
\[\large 12x^2 + 12x + 3\] \[\huge Answer \] \[\large 3(2x+1)(2x+1)\]
anonymous
  • anonymous
looks good to me you @Jhannybean ?
calculusxy
  • calculusxy
well 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)
Jhannybean
  • Jhannybean
\[\begin{align}12x^2+12x+3 &\rightarrow 4x^2+4x+1 \\&\rightarrow x^2+4x+4 \\&\rightarrow (x+2)(x+2) \\&\rightarrow \left(x+\frac{2}{4}\right)\left(x+\frac{2}{4}\right) \\&\rightarrow \left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right) \\&\rightarrow \color{red}{(2x+1)(2x+1)}\end{align} \]
Jhannybean
  • Jhannybean
For 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
  • Nnesha
if there is a GCF better to take it out |dw:1444172023392:dw|
Nnesha
  • Nnesha
|dw:1444172039740:dw|
Nnesha
  • Nnesha
\[12x^2+6x ---> 6x(2x+1)\] these are the mistakes :=)

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