Respuesta :
If it has the roots -1,2, and 4 it is:
(x+1)(x-2)(x-4)
(x^2-x-2)(x-4)
x^3-4x^2-x^2+4x-2x+8
x^3-5x^2_2x+8
Only C is equal to the equation with the roots provided and it is of degree 3
(x+1)(x-2)(x-4)
(x^2-x-2)(x-4)
x^3-4x^2-x^2+4x-2x+8
x^3-5x^2_2x+8
Only C is equal to the equation with the roots provided and it is of degree 3
Answer:
Option C) [tex]x^3-5x^2+2x+8=0[/tex]
Step-by-step explanation:
We're going to analyse every option,
A) [tex]5x^2+2x+8=0[/tex]
Observation: The degree of a polynomial is the value of the greatest exponent that can be found in all terms (after they have been simplified).
Then the degree of this polynomial is 2.
Now to find the roots of this polynomial we are going to use Bhaskara's Formula. The roots are the numbers that cause the polynomial to evaluate to zero.
Observation: Bhaskara's Formula can be applied in equations of the form [tex]ax^2+bx+c=0[/tex] where a, b and c are coefficients. The formula is:
[tex]x_{1}= \frac{-b+\sqrt{b^{2}-4ac} }{2a}[/tex]
and
[tex]x_{2}= \frac{-b-\sqrt{b^{2}-4ac} }{2a}[/tex]
This formula has two solutions. Then the polynomial has two roots.
Now applying the formula:
[tex]5x^2+2x+8=0[/tex]
a=5
b=2
c=8
[tex]x_{1}= \frac{-2+\sqrt{2^{2}-4.5.8} }{2.5}=\frac{-2+\sqrt{(-156)} }{10}[/tex]
[tex]x_{2}= \frac{-2-\sqrt{2^{2}-4.5.8} }{2.5}=\frac{-2-\sqrt{(-156)} }{10}[/tex]
As you can see we have in [tex]x_{1}[/tex] and in [tex]x_{2}[/tex][tex]\sqrt{(-156)}[/tex]
This means that the roots are conjugate complex numbers. The polynomial doesn't have real solutions.
Then this option is not the correct answer.
B) [tex](x + 2)(x - 1)(x + 4) = 0[/tex]
This polynomial is factored, then we already now the roots:
We have:
[tex](x_{0}+2)=0 \\ x_{0}=(-2)\\ (x_{1}-1)=0\\ x_{1}=1\\ (x_{2}+4)=0\\ x_{2}=(-4)[/tex]
This is because the Multiplication Property of Zero. The roots of this polynomial are:
[tex]x_{0}=(-2)\\ x_{1}=1\\ x_{2}=(-4)[/tex]
Now we need to know the degree of this polynomial. We have to resolve it by distributive multiplication.
First: [tex](x+2)(x-1)=x^2-x+2x-2=x^2+x-2[/tex]
Then,
[tex](x + 2)(x - 1)(x + 4) = (x^2+x-2)(x+4)=\\ =x^3+4x^2+x^2+4x-2x-8\\=x^3+5x^2+2x-8[/tex]
[tex]x^3+5x^2+2x-8=0[/tex]
The degree of this polynomial is 3.
Then this is not the correct answer.
C) [tex]x^3 - 5x^2 + 2x + 8 = 0[/tex]
The degree of this polynomial is 3.
To find the roots of this polynomial we have to factorize. First we're going to use the Rational root theorem.
[tex]a_{0}=8\\ a_{n}=1[/tex]
The divisors of [tex]a_{0}: 1, 2, 4, 8[/tex] and the divisors of [tex]a_{n}: 1[/tex]
Then, [tex]\pm \frac{1,2,4,8}{1}[/tex], [tex]-\frac{1}{1}[/tex] is the root of the expression.
[tex]x^3 - 5x^2 + 2x + 8 = (x+1)\frac{x^3 - 5x^2 + 2x + 8}{x+1}[/tex]
[tex]\frac{x^3 - 5x^2 + 2x + 8}{x+1}=\\x^2+\frac{-6x^2+2x+8}{x+1} =\\x^2-6x+\frac{8x+8}{x+1} =x^2-6x+8[/tex]
We're going to apply Bhaskara's Formula to the term [tex]x^2-6x+8[/tex]
[tex]x_{1}= \frac{6+\sqrt{(-6)^{2}-4.1.8} }{2.1}=\frac{6+\sqrt{(4)} }{2}[/tex]
[tex]x_{1}=4[/tex]
[tex]x_{2}= \frac{6-\sqrt{(-6)^{2}-4.1.8} }{2.1}=\frac{6-\sqrt{(4)} }{2}[/tex]
[tex]x_{2}=2[/tex]
Then the polynomial can be express in this way:
[tex]x^3 - 5x^2 + 2x + 8 = (x+1)(x-2)(x-4)[/tex]
[tex](x+1)(x-2)(x-4)=0[/tex]
The roots of the polynomial are: [tex]x_{0}=(-1)\\ x_{1}=2\\ x_{2}=(4)[/tex]
Then this is the correct answer.
D) [tex]x^4 - x^3 - 5x^2 + 2x + 8 = 0[/tex]
The degree of this polynomial is 4.
We're going to use the Rational root theorem.
[tex]a_{0}=8\\ a_{n}=1[/tex]
The divisors of [tex]a_{0}: 1, 2, 4, 8[/tex] and the divisors of [tex]a_{n}: 1[/tex]
Then, [tex]\pm \frac{1,2,4,8}{1}[/tex], [tex]+\frac{2}{1}[/tex] is the root of the expression.
[tex]x^4 - x^3 - 5x^2 + 2x + 8 =(x-2)(x^3+x^2-3x-4)[/tex]
And the expression [tex]x^3+x^2-3x-4[/tex] can't be factorized then we can't find the roots of the polynomial.
