anonymous
  • anonymous
Solve this eqution. 5x^3+20x^2-105x/5x^2-15x=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ash2326
  • ash2326
We have \[\frac{5x^3+20x^2-105x}{5x^2-15x}=0\] Let's Take 5x common from both numerator and denominator \[\frac{5x(x^2+4x-21)}{5x(x-3)}=0\] Cancel the 5x from both numerator and denominator \[\frac{\cancel {5x}(x^2+4x-21)}{\cancel {5x}(x-3)}=0\] so we have now \[\frac{x^2+4x-21}{x-3}=0\] Now let's find the factors of \[x^2+4x-21\] Compare this with \[ax^2+bx+c\] a=1 b=4 c=-21 Let's find the factors of ab=-21 such that the sum of the factors is 4 (b) 7 and -3 satisfy this 7 *-3=-21 (7)+(-3)=4 so \[4x=7x-3x\] we have now \[x^2+7x-3x-21\] take x common from the first two terms and -3 from last two terms \[x(x+7)-3(x+7)\] Take (x+7) common now \[(x-3)(x+7)\] So we have now \[\frac{(x-3)(x+7)}{x-3}=0\] Now cancel x-3 from both numerator and denominator \[\frac{\cancel{(x-3)}(x+7)}{\cancel{x-3}}=0\] we have now \[x+7=0\] so we get \[x=-7\]

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