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guapka [62]
3 years ago
3

The equation x3 + 3x2 + 5x + 15 = 0 has ______ real and _______ imaginary solution(s).

Mathematics
2 answers:
egoroff_w [7]3 years ago
7 0
The equation x3 + 3x2 + 5x + 15 = 0 has ⇒ 1 real and ⇒ 2 <span>imaginary solution(s). </span>
arlik [135]3 years ago
5 0

Answer:

Given Equation has 1 real Solution and 2 Imaginary Solution.  

Real Solution of given Equation is -3 and Imaginary solution are \sqrt{5}i\:\:and\:\:-\sqrt{5}i.

Step-by-step explanation:

Given Equation : x³ + 3x² + 5x + 15 = 0

let p(x) =  x³ + 3x² + 5x + 15

From polynomial factor theorem,

put x = -3

p(-3) = (-3)³ + 3(-3)² + 5(-3) + 15

        = -27 + 27 - 15 + 15

       = 0

So, -3 is one of the zero of the polynomial and solution of the equation.

Also, x + 3 is factor of p(x)

We get another factor by dividing p(x) by x + 3

that is x² + 5

so, to get other solution of given equation we put this factor equal to 0

x² + 5 = 0

x² = -5

x = ± √-5

x = + √5 × √-1 and   x = - √5 ×√-1

x=\sqrt{5}i\:\:and\:\:x=-\sqrt{5}i

Therefore, Given Equation has 1 real Solution and 2 Imaginary Solution.  

Real Solution of given Equation is -3 and Imaginary solution are \sqrt{5}i\:\:and\:\:-\sqrt{5}i.

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