Answer:-2 , 2+i,2-i
Step-by-step explanation:
[tex]P(x) =x^3-2x^2-3x+10[/tex]
Putting P(x) = 0
[tex]0=x^3-2x^2-3x+10[/tex]
[tex]x^3-2x^2-3x+10=0[/tex]
Let us now factorize the above equation
First we will keep different random values of x in above equation and see for which value of x, P(x) becomes 0. We will start from x=1, x=-1 , x=2 , x=-2 and so on .
When we will do that we will find that for x = -2 , P(x) becomes 0
hence x+2 is one of factor of the polynomial . In order to factorize P(x) , we divide our polynomial by ( x+2) and find the quotient
we get
[tex]x^3-2x^2-3x+1 = x^2(x+2)-4x(x+2)+5(x+2)\\x^3-2x^2-3x+1= (x+2)(x^2-4x+5)\\[/tex]
Now we factorize the quadratic polynomial above using quadratic formula which is
[tex]x=\frac{-b(+/-)\sqrt{b^2-4ac} }{2a}[/tex]
here
a = 1 , b = 4 and c = 5
putting these values in above formula we get
x= 2+i and 2-i
Hence the three roots we have are
x=-2 , 2-i and 2+i
Answer:
Option B) The roots of given polynomial are [tex]x = -2,2+i, 2-1[/tex]
Step-by-step explanation:
We are given the following polynomial:
[tex]p(x) = x^3 - 2x^2 - 3x + 10[/tex]
We have to find the zeroes of the polynomial.
A root or a zero of a polynomial are the value of x that cause the polynomial p(x) = 0.
Thus, equating the given polynomial to 0, we get:
[tex]p(x) = x^3 - 2x^2 - 3x + 10 = 0\\[/tex]
By hit and trial method we get that x = -2 is a root of the given polynomial.
[tex]p(-2) = (-2)^3 - 2(-2)^2 - 3(-2) + 10\\= -8 - 8 + 6 +10 = 0[/tex]
Thus, [tex]x+2[/tex] is a factor of the given polynomial.
Dividing the given polynomial with [tex]x+2[/tex] , we get,
[tex]p(x) = (x+2)(x^2-4x+5)[/tex]
Factorizing the factor [tex](x^2-4x+5)[/tex]
[tex](x^2-4x+5) = 0\\x^2-4x+4+1=0\\(x-2)^2+ 1 = 0\\(x-2)^2=-1\\(x-2)=\sqrt{-1}\\(x-2) = \pm i\\x = 2+i, 2-1[/tex]
Thus, the roots of given polynomial are [tex]x = -2,2+i, 2-1[/tex]
