**Introduction:**
To find a quadratic polynomial with given zeros, we need to use the fact that the sum of the zeros of a quadratic polynomial is equal to the negative coefficient of the linear term, and the product of the zeros is equal to the constant term divided by the coefficient of the quadratic term.

**Given Information:**
Zeros: -5 and 6

**Let's proceed step by step:**

**Step 1: Find the sum of the zeros:**
The sum of the zeros of a quadratic polynomial is equal to the negative coefficient of the linear term. Let's assume the quadratic polynomial as:
f(x) = ax^2 + bx + c

Sum of zeros = -b/a

Given the sum of zeros, -5 and 6, we can set up the equation:
-5 + 6 = -b/a

Simplifying the equation:
1 = -b/a

Therefore, -b = a

**Step 2: Find the product of the zeros:**
The product of the zeros of a quadratic polynomial is equal to the constant term divided by the coefficient of the quadratic term. Continuing from the previous assumption:
f(x) = ax^2 + bx + c

Product of zeros = c/a

Given the product of zeros, -5 and 6, we can set up the equation:
(-5)(6) = c/a

Simplifying the equation:
-30 = c/a

**Step 3: Substitute the values and form the quadratic polynomial:**
From Step 1, we know that -b = a. Substituting this value into the equation -30 = c/a, we get:
-30 = c/(-b)

Multiplying both sides by -b, we have:
-30(-b) = c

Simplifying the equation:
30b = c

Now we can substitute the values of a, b, and c into the quadratic polynomial f(x) = ax^2 + bx + c:
f(x) = (-b)x^2 + bx + (-b)

Simplifying further:
f(x) = bx^2 + bx - bx

Final Result:
The quadratic polynomial with zeros -5 and 6, and the sum of the zeros as -1, is:
f(x) = bx^2 + bx - bx

**Conclusion:**
We have successfully found a quadratic polynomial with zeros -5 and 6, and the sum of the zeros as -1. The polynomial is given by f(x) = bx^2 + bx - bx.

Give alpha+ beta = -5 alpha × beta = 6 general form of quadratic polynomial is p(x)= k[x² - (a+b) + ab ] put given values in general form p(x)= k [x² - (-5) +6 ] = k [x²+5 - 6] k= 1 since k is constant therefore required polynomial is x² + 5 + 6