Question

Find a quadratic polynomial whose zeroes are 3,-2.

Answer 1

Hello 

If the roots are a and b, the equation of the function might be (x-a)(x-b) or most generally c(x-a)(x-b) 

so you can have 
[x-(2+√3)]*[x-(2-√3)] 
= (x-2-√3)(x-2+√3) 
= x² -2x +x√3 -2x +4 -2√3 -x√3 +2√3 -(√3)² 
= x² - 4x + 4 - 3 
= x² - 4x + 1 

Check : 

Δ = b² - 4ac = 16 - 4(1)(1) = 12 
√Δ = √12 = 2√3 

x1 = (4 + 2√3)/2 = 2 + √3 
x2 = (4 - 2√3)/2 = 2 - √3 

Hope it helps