Question

Write the zeroes of the polynomial x square -x -6

Answer 1

Answer:

-2, 3

Step-by-step explanation:

To obtain the zeroes or roots of the polynomial, we need to equate the given equation to zero i.e, x^2 - x - 6 = 0   (x^2 denotes x square)

Now to solve the equation we need to find the prime factors of 6(coefficient of constant term). There are only two prime factors for 6 that are 2 and 3.

Now we need to expand the equation using the prime factors of 6 such that the product of prime factors is equal to 6 and their sum is equal to the coefficient of x. (Note that here we use negative numbers as factors at times to solve the problems but in general there does not exist negative prime factors for a number.)

=> x^2 - x - 6 = 0

=> x^2 - 3x + 2x - 6 = 0

Now we need to obtain the common terms between them.

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

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

=> x+2 = 0           => x-3 = 0

=> x = -2              => x = 3

So our zeroes for the given equation is -2 and 3.

Answer 2

Answer- The above question is from the chapter 'Polynomials'.

Let's know about polynomials first.

Polynomial- It is an algebraic expression involving use of variables and constants.

p(x)- It is used to denote a polynomial. It is read is 'Polynomial in x'.

Polynomials can be classified on two basis:

1) Number of terms

E.g.- Polynomial with one term is called monomial.

Two terms- binomial

Three terms- trinomial

2) Power of variable

E.g.- Polynomial with degree 1 is called linear polynomial.

degree 2- quadratic polynomial

degree 3- cubic polynomial

degree 4- bi-quadratic polynomial

Quadratic polynomial- A polynomial whose highest power of variable is 2 is called a quadratic polynomial.

Examples:

1) x² + 2x + 2 = 2

2) 2x² + 4x + 1 = 0

Relationship between zeroes and coefficients of a quadratic equation:

Let ax² + bx + c = 0 be any quadratic equation.

Let α and β be its zeroes.

Sum of zeroes i.e α and β= -b/a

Product of zeroes i.e αβ= c/a

Discriminant (D) = b² - 4ac

Given question: Write the zeroes of the polynomial x² - x - 6.

Solution: Let p(x) = x² - x - 6

= x² - 3x + 2x - 6

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

= (x - 3) (x + 2)

Let p(x) be 0.

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

Either x - 3 = 0

x = 3

Or x + 2 = 0

x = -2

∴ Zeroes of p(x) =  x² - x - 6 are 3 and -2.