Find the zeros of the quadratic polynomial x2-2x-8 and verify the relationship. | the number of zeros of the polynomial x2-2x-8 will be? | x2 – 2x – 8 = 0

x2-2x-8=0 quadratic formula

x2-2x-8 factorise

the zeros of x2-2x-8 are

x2-2x-8 class 10

## x2-2x-8 = 0

We are going to find the zeros of the quadratic polynomial x^{2} – 2x – 8 = 0.

After seeing this equation we find that we need two numbers in such a way that the multiplication of these two numbers should be equal to -8. And the addition of these two numbers should be equal to -2.

We will find these two numbers by factorising 8.

8 = 2x2x2 = 4 x 2

So, these two numbers are -4 and 2. Because -4×2 = -8 And -4 + 2 = -2

## x^{2} – 2x – 8 = 0

First Take LHS (Left Hand Side)

x^{2} – 2x – 8

Factorise x^{2} – 2x – 8

= x^{2} -4x + 2x – 8

= x(x-4) +2(x-4)

= (x+2)(x-4)

### The zeros of x2-2x-8 are

(x+2) = 0

(x-4) = 0

x = -2

x= 4

the number of zeros of the polynomial x2-2x-8 will be x = -2 and x = 4

Also, Read:

Find the Factorization of 3x^{2} -8x + 5 = 0

### x2-2x-8 =0 Quadratic Formula Verify The Relationship

x^{2} – 2x – 8 = 0 ( equ.____1)

L.H.S.

x^{2} – 2x – 8

= x^{2} – 4x + 2x – 8

= x(x-4) +2(x-4)

= (x-4)(x+2)

(x-4)(x+2) = 0

x = 4

x = -2

So, we find two value of x.

Let’s take 1^{st} value and put in equ. 1

x^{2} – 2x – 8 = 0

putting x = 4

4^{2} – 2 * 4 – 8 = 0

16 – 8 – 8 = 0

16 -16 = 0

0 = 0

Now Putting the value of x = -2 in equ. 1

x^{2} – 2x – 8 = 0

putting x = -2

(-2)^{2} – 2 * (-2) – 8 = 0

4 + 4 – 8 = 0

8 – 8 = 0

0 = 0

Hence from both value of x we find

LHS = RHS

Hence we prove this relation.

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