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nisha sharma added a new attachment: kindly resolve.... WhatsApp Image 2017-06-14 at 11.17.09 AM.jpeg
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  • June 14, 2017 at 6:17am
  • Little Sky:[deqn]f(z)=\text{quotient}\times\text{divisor}+\text{remainder}[/deqn] For case [ieqn]f(z)=az^3+4z^2+3z-4=\text{quotient_1}\times(z-3)+\text{common remainder}[/ieqn] For case [ieqn]f(z)=z^3-4z-a=\text{quotient_2}\times(z-3)+\text{common remainder}[/ieqn] You should observe that for both cases, when z = 3, you are left with the common remainder. For 1st case, [ieqn]f(3)=a(3)^3+4(3)^2+3(3)-4=27a+36+9-4=27a+41[/ieqn] For 2nd case, [ieqn]f(3)=3^3-4(3)-a=15-a[/ieqn] By equating the remainder 27a + 41 and 15 - a, you should get a = -26/28
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    • June 14, 2017 at 7:16am
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