Question

find the zero of polynomial p(x)=6²-7x-3 and show the relation of the zeroes and the coefficient​

Answer 1

Solution :

Here, we'll use the concept of splitting the middle term in finding out of zeroes of the given quadratic polynomial. In this process we'll split the middle term into two parts whose sum will be equal to the coefficient of x and product will be equal to constant term.

Here

A polynomial is given as p(x) 6x² - 7x - 3.

Now we'll compare the polynomial 6x² - 7x - 3 with ax² + bx + c

In this polynomial :

• a = 6 (coefficient of x²)
• b = -7 (coefficient of x)
• c = -3 (constant term)

Now, we'll find the zeroes of the polynomial.

=> 6x² - 7x - 3

=> 6x² - 9x + 2x - 3

=> 3x (2x - 3) + 1 (2x - 3)

=> (3x + 1) (2x - 3)

=> x = -1/3 , 3/2

∴ Zeroes of the polynomial p(x) 6x² - 7x - 3 are -1/3 and 3/2 respectively.

Now,

Sum of zeroes :

=> -1/3 + 3/2

=> (-2+9)/6

=> 7/6 = -b/a

Product of zeroes :

=> -1/3 × 3/2

=> -1/2 (-3/6) = c/a

Hence, verified !