Applying the factor theorem, it is found that the roots of the equation are:
[tex]x = 1, x = -4, x = 2 + 5i, x = 2 - 5i[/tex]
The factor theorem states that if [tex]x_1, x_2, ..., x_n[/tex] are roots of a polynomial, it can be written as:
[tex](x - x_1)(x - x_2)...(x - x_n)[/tex]
In this problem:
[tex](x^2 + 3x - 4)(x^2 - 4x + 29) = 0[/tex]
Thus, the roots are the values of x for which either:
[tex]x^2 + 3x - 4 = 0[/tex]
Or
[tex]x^2 - 4x + 29 = 0[/tex]
First, [tex]x^2 + 3x - 4 = 0[/tex]
Which is a quadratic equation with [tex]a = 1, b = 3, c = -4[/tex], thus:
[tex]\Delta = 3^{2} - 4(1)(-4) = 25[/tex]
[tex]x_{1} = \frac{-3 + \sqrt{25}}{2} = 1[/tex]
[tex]x_{2} = \frac{-3 - \sqrt{25}}{2} = -4[/tex]
Thus, [tex]x = 1[/tex] and [tex]x = -4[/tex] are roots.
Then, we solve [tex]x^2 - 4x + 29 = 0[/tex].
The coefficients are [tex]a = 1, b = -4, c = 29[/tex], so:
[tex]\Delta = (-4)^{2} - 4(1)(29) = -100[/tex]
[tex]x_{1} = \frac{-(-4) + \sqrt{100}}{2} = 2 + 5i[/tex]
[tex]x_{2} = \frac{-(-4) - \sqrt{100}}{2} = 2 - 5i[/tex]
Thus, [tex]x = 2 + 5i[/tex] and [tex]x = 2 - 5i[/tex] are also roots.
A similar problem is given at https://brainly.com/question/24380382
