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Kitty [74]
3 years ago
15

A cubic function with roots at 2, 0 and 3 containing the point at (5,-6)

Mathematics
1 answer:
natulia [17]3 years ago
5 0

Answer:

Cubic equations and the nature of their roots

Just as a quadratic equation may have two real roots, so a cubic equation has possibly three. But unlike a quadratic equation which may have no real solution, a cubic equation always has at least one real root.

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Make a table of ordered pairs for the equation.
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Graph the line using the slope and y-intercept, or two points.

Slope: -1/3

Y-Intercept: (0,1)

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How many tenths equal to a one
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solve and check: (1)/(x 3) = (x 10)/(x − 2) from least to greatest, the solutions are x = ? and x = ?
SCORPION-xisa [38]
If you would like to solve 1/(x + 3) = (x + 10)/(x - 2), you can do this using the following steps:

<span>1/(x + 3) = (x + 10)/(x - 2)
</span>1 * (x - 2) = (x + 10) * (x + 3)
x - 2 = x^2 + 3x + 10x + 30
0 = x^2 + 13x + 30 - x + 2
0 =  x^2 + 12x + 32
0 = (x + 8) * (x + 4)
1. x = -8
2. x = -4

The correct result would be x = -8 and x = -4.
5 0
3 years ago
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A) Find all zeros of P, real and complex. (Enter your answers as a comma-separated list.
lara31 [8.8K]

Answer:

steps below

Step-by-step explanation:

x⁶ - 7x³ - 8 = (x⁶ + x³) - (8x³ + 8)

                  = x³ (x³+1)-8(x³+1)

                  = (x³ - 2³)(x³ + 1)

                  = (x-2)(x²+2x+4)(x+1)(x²-x+1)

(x-2)(x²+2x+4)(x+1)(x²-x+1) = 0

<u>x= -1 or x = 2</u>   ... roots

or    x²+2x+4=0   or    x²-x+1=0  

x²+2x+4=0

x = (-2±√4-16)/2 = <u>-1 ± √3 i</u>   ... complex roots

x²-x+1=0

x = (1±√1-4)/2 = <u>1/2 ± (√3)i / 2</u>   ... complex roots

<u />

b) P(x) = x⁶ - 7x³ - 8

           = <u>(x-2)(x²+2x+4)(x+1)(x²-x+1)</u>

           = (x+1)(x-2)(x-(-1 + √3 i))(x-(-1 - √3 i)(x-(1/2 + (√3)i / 2))(x-(1/2 - (√3)i / 2))

6 0
3 years ago
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