Answer: x1=1 x2=-2 and x3=2
Step-by-step explanation:
1st x1=1 is 1 of the roots , so
F(1)=1-1-4+4=0 - true
So lets divide x^3-x^2-4x+4 by (x-x1), i.e (x^3-x^2-4x+4) /(x-1)=(x^2-4)
x^2-4 can be factorized as (x-2)*(x+2)
So x^3-x^2-4x+4=(x-1)*(x^2-4)=(x-1)(x-2)*(x+2)
So there are 3 dofferent roots:
x1=1 x2=-2 and x3=2
- If x = 5 is a zero of the given polynomial, then,
- (x - 5) is a factor of the polynomial. [Since, x = 5 or, x - 5 = 0]
- Now, divide the polynomial with (x - 5) using long division method. (See the picture)
- We get (x^2 - x - 6) as the quotient.
- Now, factorise the above polynomial:
- (x^2 - x - 6)
- = (x^2 - 3x + 2x - 6)
- = x(x - 3) + 2(x - 3)
- = (x - 3)(x + 2)
- Therefore, x^3 − 6x^2 − x + 30 = (x − 5)(x − 3)(x + 2)
<u>Answer</u><u>:</u>
<u>B.</u><u>(x − 5)(x − 3)(x + 2)</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
The correct answer should be B because it does cross the y-axis at -5 and it does decres from x = -10 till x = -2 and then does remain constant till x=10