PART B: CHOOSE 2 From this Part! State the number of complex roots and the possible number of real and imaginary roots for each equation. Then find all roots.

Question

1) f (x) = x 3 + 2x 2 - 8x 3) f (x) = x 3 + 2x 2 - 3x

2)f (x) = x 3 + 5x 2 + 6x 4) f (x) = x 3 - 9x 2 + 20x

Answer 1

Hi,

1) f (x) = x 3 + 2x 2 - 8x first factor of x ------> x( x^2 +2x -8) = 0 -------> x = 0 and x^2 +2x -8 = 0

Now, you find b^2 -4ac that in this case will be = 2^2 -4*1*(-8) = 4 + 32 = 36 it means this equation has 2 different real roots and will be (-b ±√b^2 - 4ac) / 2a ------> x = 2 , -4 as result #1 has 3 real roots such as 0,2,-4

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2)f (x) = x 3 + 5x 2 + 6x completely the same

x 3 + 5x 2 + 6x = 0

x( x^2 + 5x +6 ) = x ( x + 2 ) ( x + 3 ) = 0 all roots will be 0, -2, -3

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3) f (x) = x 3 + 2x 2 - 3x the same

factor of x : x( x^2 +2x -3 ) = x ( x +3 ) ( x - 1 ) = 0 all roots will be 0, -3, 1

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4) f (x) = x 3 - 9x 2 + 20x the same

x( x^2 -9x + 20 ) = x (x - 4 ) ( x - 5 ) =0 all real roots will be 0, 4, 5

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I hope it is useful,

Minoo