Question

the zeroes of polynomial of 6x^2-3x-7x​

Answer 1

$\huge\mathfrak{Answer:}$

Correct Question:

Find the zeroes of the polynomial 6x² - 7x - 3.

Given:

• We have been given a quadratic polynomial 6x² - 7x - 3.

To Find:

• We need to find the zeroes of this polynomial.

Solution:

We can find the zeroes of this quadratic polynomial by the method of splitting the middle term.

We need to find two such numbers whose sum is -7 and product is -18.

Two such numbers are -9 and 2.

Substituting the values, we have

6x² -9x + 2x - 3 = 0

=> 3x(2x - 3) + 1(2x - 3) = 0

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

Either (2x - 3) = 0 or (3x + 1) = 0.

When (2x - 3) = 0

=> 2x = 3

=> x = 3/2

When (3x + 1) = 0

=> 3x = -1

=> x = -1/3

Hence, the two zeroes of this polynomial are 3/2 and -1/3.

Answer 2

QUESTION:

The zeroes of polynomial of $\rm{6{x}^{2}-3-7x}$

ANSWER:

The zeros of the polynomials are : $\rm{\frac{3}{2} and - \frac{1}{3}}$

GIVEN:

The given polynomial is :$\rm{6{x}^{2}-3-7x}$

TO FIND:

The roots of the polynomial

SOLUTION:

The given polynomial is :$\rm{6{x}^{2}-3-7x}$

$6 {x}^{2} - 3 - 7x$

$= \: 6 {x}^{2} - 7x - 3$

$= \: 6 {x}^{2} - 9x + 2x - 3$

$= \: 3x(2x - 3) + 1(2x - 3)$

$= \: (2x - 3)(3x + 1)$

$x = \frac{3}{2} , \: x = - \frac{1}{3}$

⛬ The zeros of the polynomials are : $\rm\large{\frac{3}{2} \ and - \frac{1}{3}}$