To find the relationship between the zeroes and the coefficients of the polynomial x2−2x−8x^2 – 2x – 8, follow these steps:

Step 1: Identify the given quadratic polynomial

The given polynomial is:

x2−2x−8x^2 – 2x – 8

Here, the general form of a quadratic polynomial is:

ax2+bx+cax^2 + bx + c

Comparing, we get:

• a=1a = 1
• b=−2b = -2
• c=−8c = -8

Step 2: Find the zeroes of the polynomial

To find the zeroes, solve:

x2−2x−8=0x^2 – 2x – 8 = 0

Factorizing,

(x−4)(x+2)=0(x – 4)(x + 2) = 0

Setting each factor equal to zero,

x−4=0 or x+2=0 

So, the zeroes of the polynomial are 4 and -2.

Step 3: Verify the Relationship Between Zeroes and Coefficients

According to Vieta’s formulas:

1. Sum of the zeroes:

   α+β=−baSubstituting values,

   4+(−2)=−−21=2Matches the formula.

2. Product of the zeroes:

   α⋅β=caSubstituting values,

   4×(−2)=−81=−84 \times (-2) = \frac{-8}{1} = -8Matches the formula.

Conclusion

For the quadratic polynomial x2−2x−8x^2 – 2x – 8, the relationship between the zeroes and the coefficients is:

• Sum of zeroes (α+β)=−ba=2(\alpha + \beta) = -\frac{b}{a} = 2
• Product of zeroes (α⋅β)=ca=−8(\alpha \cdot \beta) = \frac{c}{a} = -8