if alpha and beta are the zero's of quadratic polynomial f(x) =x*x-x-2 find a polynomial who's zeros are 2 alpha +1 and 2 bita + 1 ?
You can factorize your given polynomial as follows:
x^2 - x - 2 = 0
=> (x + 1)(x - 2) = 0
The roots are thus x = 2, x = -1
Alpha = 2, beta = -1
Coming to the next part, (2alpha + 1) = 5, (2beta + 1) = -1
You can thus use these roots to form the following polynomial:
(x - 5)(x + 1) = 0
=> x^2 - 4x - 5 = 0
To find a polynomial with zeros 2α + 1 and 2β + 1, we can follow these steps:
Step 1: Recall that if α and β are zeros of the quadratic polynomial f(x) = x^2 - x - 2, then we can write the factors as follows:
(x - α)(x - β) = 0
Step 2: Substitute 2α + 1 and 2β + 1 into the factors:
[(x - (2α + 1))][(x - (2β + 1))] = 0
Step 3: Simplify the expression inside the brackets:
[(x - 2α - 1)][(x - 2β - 1)] = 0
Step 4: Expand the equation by multiplying the brackets:
x^2 - 2αx - x + 2α - 1 - 2βx + 2αβ + β - 1 = 0
Step 5: Simplify the terms:
x^2 - (2α + 2β + 2)x + (2αβ + 2α + β - 2) = 0
So, the polynomial with zeros 2α + 1 and 2β + 1 can be written as:
P(x) = x^2 - (2α + 2β + 2)x + (2αβ + 2α + β - 2)
To find a polynomial with zeros 2α + 1 and 2β + 1, we can make use of the fact that if α and β are the zeros of a quadratic polynomial f(x) = x^2 - x - 2, then the polynomial can be factored as (x - α)(x - β).
To obtain a polynomial with the desired zeros, we substitute (x - (2α + 1))(x - (2β + 1)) into the equation.
Let's expand the expression:
(x - (2α + 1))(x - (2β + 1))
= x^2 - x(2β + 1) - (2α + 1)x + (2α + 1)(2β + 1)
= x^2 - 2βx - x - 2αx + 2α - 2β - x + 2αβ + 2β + 1
= x^2 - (3α + 3β + 3)x + (2α - 2β + 2αβ + 1)
Therefore, a polynomial with zeros 2α + 1 and 2β + 1 is given by:
f(x) = x^2 - (3α + 3β + 3)x + (2α - 2β + 2αβ + 1)