anonymous
  • anonymous
Help me understand quadratic equations?? Pic in comments
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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AccessDenied
  • AccessDenied
Is there anything in particular that you are unsure of? Or is it the entire concept?
anonymous
  • anonymous
entire concept, ive had a lot on my mind lately and havent been paying attention in class... i =ve been thinking of other things... /:

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AccessDenied
  • AccessDenied
Hmm... well, in essence, we're taking the process of multiplying two binomials in reverse e.g. here's a random example I made (try to bear with explanation, its kinda long) (2x + 3)(x - 4) 2x(x - 4) + 3(x - 4) 2x^2 - 8x + 3x - 12 2x^2 - 5x - 12 The general process of factoring the quadratic is like this. We multiply together the leading coefficient to the constant term (in this case, 2 and -12). 2 * -12 = -24. We are looking for factors of -24 that add up to -8. If they add up to a negative, then the larger number is negative of two factors. -24 * 1 -> -24 + 1 = -23 -12 * 2 -> -12 + 2 = -10 -8 * 3 -> -8 + 3 = -5 < There it is. We will rewrite the expression using these as the linear term's coefficients (since rewriting -5x as -8x + 3x is still correct): 2x^2 - 8x + 3x - 12 <- If we look at the original multiplication we did, we can see that this came up as well Then we factor by grouping those first two terms and last two terms together (2x^2 - 8x) + (3x - 12) if we factor out the greatest common factor, then we should get a common factor. 2x(x - 4) + 3(x - 4) Then we factor out the (x-4) (2x + 3)(x - 4) This looks like the expression we had originally! This is probably a lot to take in at once... We'll try to work through some more examples and maybe that will help. :)
AccessDenied
  • AccessDenied
although, i hope that all makes sense. not sure if you did factoring of any other quadratics like just "x^2 + 4x + 4," since your example is mostly factoring those ones with the non-1 leading coefficient.
AccessDenied
  • AccessDenied
\[ \color{blue}ax^2 + \color{goldenrod}bx + \color{red}c\\ We~take~\color{blue}a~times~\color{red}c~and~determine~which~factors~of~\color{blue}a\color{red}c~add~up~to~\color{goldenrod}b.\\ We~rewrite~\color{goldenrod}bx~as~\color{olive}{b_1}x~+~\color{olive}{b_2}x.\\ \color{blue}ax^2 + \color{olive}{b_1}x + \color{olive}{b_2}x + \color{red}c\\\\ We~group~together~\color{blue}ax^2~+~\color{olive}{b_1}x~and~\color{olive}{b_2}x~+~\color{red}c~and~find~the~greatest~common~factor.\\ We~factor~out~the~greatest~common~factor~of~each\\~~~group~to~get~a~common~binomial~between~the~two.\\ Then,~we~factor~out~the~common~binomial~and~we~are~done!\\ \]

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