Question

sum of zeroes and product of zeros ​

Answer 1

x^2-2x-8

x^2-4x+2x-8

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

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

x+2=0 , x-4=0

x=-2 , x=4

α=-2. ,β=4

now realtion where a=1,b=-2,c=-8

•α+β=-b/a

-2+4=-(-2)/1

2=2

verified

•α*β=c/a

4*-2=-8/1

-8=-8

verified

Answer 2

x² - 2x - 8

Here, we've to find the zeros of the polynomial. It means that, we've to find that numbers which can be substituted in the place of x so that the result will be zero.

We've to factorise it. We can do it with many methods but here fpr pur simplicity I'll do, it with splitting the middle term method.

Step 1: Write the polynomial correctly. Sometimes, the polynomial is not given in the standard form i.e., ax^2 + bx + c = 0, so, we've to rearrange the polynomial without changing their signs.

Here, the polynomial x² - 2x - 8 is in standard form. So, we don't wanna to write it again or rearrange it.

Step 2: Multiply the co-efficient of x² and the constant term.

Here, the co-efficient of x² is 1 and the constant term is - 8.

Multiplying 1 and - 8 = - 8

Step 3: Find the two factors of - 8 which would gives us the sum as the co-efficient of x.

Factors of - 8 = $\pm 1, \pm 2, \pm 4, \pm 8$

So, the two factors of - 8 which would give us the sum as - 2 are - 4 and 2.

Step 4: Substituting - 4 and 2 in the place of - 2 along with their respective variable.

x² - 2x - 8

x² - 4x + 2x - 8

Step 4: Group the terms.

(x² - 4x) + (2x - 8)

Step 5: Take common term outside.

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

Step 6: Now you'll observe that, both the terms inside the bracket will be same as the other.

Write the common terms first in side the bracket and uncommon terms inside another bracket to be written after the common term.

(x - 4)(x + 2)

Therefore, (x - 4) and (x + 2) are the factors of x^2 - 2x - 8.

As we have to find zeros, then the polynomial x² - 2x - 8 should be equal to zero. If the polynomial x² - 2x - 8 is equal to zero, then it's factors (x - 4) and (x + 2) should also be equal to zero.

Step 7: Equating the factors (individually) with zero.

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

(x - 4) = 0

x - 4 = 0

$\boxed{\bold{\mathsf{x = 4}}}$

(x + 2) = 0

x + 2 = 0

$\boxed{\bold{\mathsf{x = - 2}}}$

Therefore, x = 4 or x = 2 are two zero of the polynomial x^2 - 2x - 8.

To verify the relationship between the zeros and their co - efficients.

The following are the relationships between the co-efficients and the zeros of a quadratic polynomial.

Let the zeros be $\alpha\:\&\: \beta$ respectively.

$\boxed{\alpha + \beta = \frac{-b}{a}}$

$\boxed{\alpha \times \beta = \frac{c}{a}}$

(i) $\alpha + \beta = \frac{-b}{a}$

$4 + (-2) = \frac{-(-2)}{1}$

2 = 2

$\fbox{\bold{\mathsf{L.H.S = R.H.S}}}$

(ii) $\alpha \times \beta = \frac{c}{a}$

$4 \times (-2) = \frac{-8}{1}$

- 8 = - 8

$\fbox{\bold{\mathsf{L.H.S = R.H.S}}}$

Therefore, the relationship between the zeros and the co-efficient of the polynomial is hence verified.

Therefore, the zeros of the polynomial x² - 2x - 8 are 4 and - 2.