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Determine the number of possible imaginary (complex) solutions. Provide multiple answers separated by commas.

a) [tex]x^4-7x^3-x^2+67x-60[/tex]

Options: 0, 2, 4

b) [tex]-4x+4[/tex]

Options: 0

c) [tex]x^2+2x-3[/tex]


Sagot :

Let's determine the number of possible imaginary (complex) solutions for each given polynomial.

### Part (a): [tex]\( x^4 - 7x^3 - x^2 + 67x - 60 = 0 \)[/tex]

To find the roots of the polynomial [tex]\( x^4 - 7x^3 - x^2 + 67x - 60 \)[/tex], apply the Fundamental Theorem of Algebra, which states that a polynomial of degree [tex]\( n \)[/tex] has exactly [tex]\( n \)[/tex] roots, considering their multiplicity and including complex roots.

1. The polynomial is of degree 4, so there will be 4 roots in total.
2. Polynomials with real coefficients have roots that appear in conjugate pairs if they are complex.

To determine the number of complex (imaginary) roots:
- Check if the polynomial can be factored easily. For polynomials with real coefficients, if a complex number is a root, then its conjugate is also a root.

From the provided information, we know:
- There are 0 imaginary (complex) roots.

### Part (b): [tex]\( -4x + 4 = 0 \)[/tex]

For the linear equation [tex]\( -4x + 4 = 0 \)[/tex]:

1. This is a linear polynomial of degree 1.
2. A linear polynomial has exactly one root.
3. Since the coefficient of [tex]\( x \)[/tex] and the constant term are real numbers, there cannot be any complex roots.

From the provided information, we know:
- There are 0 imaginary (complex) roots.

### Part (c): [tex]\( x^2 + 2x - 3 = 0 \)[/tex]

To solve the quadratic equation [tex]\( x^2 + 2x - 3 = 0 \)[/tex]:

1. This is a quadratic polynomial of degree 2, so it can have up to 2 roots.
2. Use the quadratic formula to determine the roots: [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].

Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -3 \)[/tex].

The discriminant [tex]\( \Delta \)[/tex] is given by [tex]\( \Delta = b^2 - 4ac \)[/tex]. Plug in the values:
[tex]\[ \Delta = 2^2 - 4(1)(-3) = 4 + 12 = 16 \][/tex]

Since the discriminant is positive ([tex]\( \Delta = 16 \)[/tex]), the quadratic equation has two distinct real roots.

From the provided information, we know:
- There are 0 imaginary (complex) roots.

In summary:

- For [tex]\( x^4 - 7x^3 - x^2 + 67x - 60 = 0 \)[/tex]:
- Number of imaginary (complex) roots: 0

- For [tex]\( -4x + 4 = 0 \)[/tex]:
- Number of imaginary (complex) roots: 0

- For [tex]\( x^2 + 2x - 3 = 0 \)[/tex]:
- Number of imaginary (complex) roots: 0