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Question: If $\alpha$ and $\beta$ are the zeros of the polynomial f(x) = $6x^2-3-7x$ then $(\alpha + 1) (\beta...

If α\alpha and β\beta are the zeros of the polynomial f(x) = 6x2−3−7x6x^2-3-7x then (α+1)(β+1)(\alpha + 1) (\beta + 1) is equal to -

A

52\frac{5}{2}

B

53\frac{5}{3}

C

25\frac{2}{5}

D

35\frac{3}{5}

Answer

53\frac{5}{3}

Explanation

Solution

The given polynomial is f(x)=6x2−7x−3f(x) = 6x^2 - 7x - 3. For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of roots is α+β=−b/a\alpha + \beta = -b/a and the product of roots is αβ=c/a\alpha\beta = c/a. Here, a=6a=6, b=−7b=-7, and c=−3c=-3. So, α+β=−(−7)/6=7/6\alpha + \beta = -(-7)/6 = 7/6. And αβ=−3/6=−1/2\alpha\beta = -3/6 = -1/2. We need to find (α+1)(β+1)(\alpha + 1)(\beta + 1). Expanding this expression: (α+1)(β+1)=αβ+α+β+1(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1. Substituting the values: (α+1)(β+1)=(−1/2)+(7/6)+1(\alpha + 1)(\beta + 1) = (-1/2) + (7/6) + 1. Simplifying with a common denominator of 6: =−3/6+7/6+6/6=(−3+7+6)/6=10/6=5/3= -3/6 + 7/6 + 6/6 = (-3 + 7 + 6) / 6 = 10/6 = 5/3.