Finding the zeroes of a quadratic polynomial

To find the zeroes of a quadratic polynomial, we can use the quadratic formula or factorization method. The quadratic formula states that for a quadratic equation of the form ax2 + bx + c = 0, the roots are given by:

x = (-b ± √(b2 - 4ac)) / 2a

Alternatively, we can use the factorization method to find the roots. If we can factor the quadratic polynomial into the form (x - r)(x - s), then the roots are given by r and s.

Applying the formula to the given polynomial

The given quadratic polynomial is x2 + 3x - (a - 2)(a - 1). To find the roots, we can apply the quadratic formula:

x = (-3 ± √(32 - 4(1)(-(a-2)(a-1)))) / 2(1)
= (-3 ± √(9 + 4a2 - 4a + 4)) / 2
= (-3 ± √(4a2 - 4a + 13)) / 2

Thus, the roots are (-3 + √(4a2 - 4a + 13)) / 2 and (-3 - √(4a2 - 4a + 13)) / 2.

Simplifying the expression

We can simplify the expression by noticing that the coefficients of the quadratic polynomial are symmetrical. That is, the coefficient of x2 is 1, the coefficient of x is 3, and the constant term is -(a-2)(a-1), which is also symmetrical. Therefore, the roots of the polynomial must be symmetrical as well.

In other words, if r is a root of the polynomial, then so is 3 - r. To see this, we can use the factorization method. If (x - r) is a factor of the polynomial, then so is (x - (3-r)), since:

(x - (3-r)) = -x + 3 + r
(x - r) + (-x + 3 + r) = 3

Therefore, the roots of the polynomial must be of the form (-3 ± k) / 2, where k is some expression involving a.

Using this form, we can simplify the expression for the roots as follows:

(-3 + k) / 2 = (-3 + √(4a2 - 4a + 13)) / 2
k = √(4a2 - 4a + 13) + 3

(-3 - k) / 2 = (-3 - √(4a2 - 4a + 13)) / 2
k = -√(4a2 - 4a + 13) + 3

Therefore, the roots are (-3 + √(4a2 - 4a + 13)) / 2 and (-3 - √(4a2 - 4a + 13)) / 2, which are equivalent to -(a-2) and (a-1), respectively.

Conclusion

The zeroes of the quadratic polynomial x2 + 3x - (a-2)(a-1) are -(a-2) and (a-