State how many imaginary and real zeros the function has. f(x) = x³ + 5x² - 28x - 32

Solution:

It is given that,

f(x) = x³ + 5x² - 28x - 32

Now we have to find the number of imaginary and real zeros the function has.

Since x = -1 is the zero of the given polynomial, therefore x + 1 is the factor of f(x) = x³ + 5x² - 28x - 32.

We need to perform factorization of  f(x) = x³ + 5x² - 28x - 32 

By using long division method ;

Above equation can be written as,

= (x + 1) (x² + 4x -32),

Now we have to perform factorization of (x² + 4x -32),

by using middle term factorisation,

= (x + 1) (x² + 8x - 4x -32)

= (x + 1) (x(x + 8) - 4 (x + 8))

= (x + 1) (x - 4) (x + 8)

From the solution, we clearly see that there were three real zeros in the given function.

Therefore, the function has 0 imaginary and 3 real zeros.

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State how many imaginary and real zeros the function has. f(x) = x³ + 5x² - 28x – 32

Summary:

The function f(x) = x³ + 5x² - 28x - 32 has 0 imaginary and 3 real zeros.