Answer :

Zeroes of the polynomial are the values of the variable of the polynomial when the polynomial is put equal to zero.

Let p(x) be a polynomial with any number of terms any number of degree.

Now, zeroes of the polynomial will be the values of x at which p(x) = 0.

If p(x) = ax^{2 }+ bx + c is a quadratic polynomial (highest power is equal to 2) and its roots are α and β, then

Sum of the roots = α + β = - b/a

Product of roots = αβ = c/a

**(i)** p(x) = x^{2 }- 2x - 8

So, the zeroes will be the values of x at which p(x) = 0

Therefore,

⇒ x^{2} - 4x + 2x - 8 = 0

(We will factorize 2 such that the product of the factors is equal to 8 and difference is equal to 2)

⇒ x(x - 4) + 2(x - 4) = 0

= (x - 4)(x + 2)

The value of x^{2 }- 2x - 8 is zero when *x* − 4 = 0 or *x* + 2 = 0,

i.e, *x* = 4 or *x* = −2

Therefore, The zeroes of x^{2 }- 2x - 8 are 4 and −2.

Sum of zeroes = 4 + (-2) = 2

Hence, it is verified that,

Product of zeroes =

Hence, it is verified that,

**(ii)** 4s^{2 }- 4s + 1

= (2s)^{2} - 2(2s)1 + 1^{2}

As, we know (a - b)^{2} = a^{2} - 2ab + b^{2}, the above equation can be written as

= (2s - 1)^{2}

The value of 4*s*^{2} − 4*s* + 1 is zero when 2*s* − 1 = 0, when, s = 1/2 , 1/2.

Therefore, the zeroes of 4*s*^{2} − 4*s* + 1 are and .

Sum of zeroes =

Product of zeroes =

Hence Verified.

**(iii)**** **6x^{2 }- 3 - 7x

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

(We will factorize 7 such that the product of the factors is equal to 18 and the difference is equal to - 7)

= 6x^{2} + 2x - 9x - 3

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

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

The value of 6*x*^{2} − 3 − 7*x* is zero when 3*x* + 1 = 0 or 2*x* − 3 = 0,

i.e.

Therefore, the zeroes of 6*x*^{2} − 3 − 7*x* are .

Sum of zeroes =

Product of zeroes =

Hence, verified.

**(iv)** 4u^{2 }+ 8u

= 4u^{2 }+ 8u + 0

= 4u (u + 2)

The value of 4*u*^{2} + 8*u* is zero when 4*u* = 0 or *u* + 2 = 0,

i.e., *u* = 0 or *u* = −2

Therefore, the zeroes of 4*u*^{2} + 8*u* are 0 and −2.

Sum of zeroes =

Product of zeroes =

**(v)** t^{2} – 15

= t^{2} – (√15)^{2}

= (t - √15)(t + √15) [As, x^{2 }- y^{2} = (x - y)(x + y)]

The value of *t*^{2} − 15 is zero when (t - √15) = 0 or (t + √15) = 0,

i.e., when t = √15 or t = -√15

Therefore, the zeroes of *t*^{2} − 15 are √15 and -√15.

Sum of zeroes =

Product of zeroes =

Hence verified.

**(vi)** 3x^{2} – x – 4

(We will factorize 1 in such a way that the product of factors is equal to 12 and the difference is equal to 1)

= 3x^{2} - 4x + 3x - 4

= x(3x - 4) + 1(3x - 4)

= (3x – 4 )(x + 1)

The value of 3*x*^{2} − *x* − 4 is zero when 3*x* − 4 = 0 or *x* + 1 = 0,

when

Therefore, the zeroes of 3*x*^{2} − *x* − 4 are and -1

Sum of zeroes =

Product of zeroes =

Hence, verified.

Rate this question :

In the given figuRS Aggarwal - Mathematics

Find all the zeroMathematics - Board Papers