Question

verify whether the following are zeroes of the polynomial indicated against them.​

Answer 1

Answer :

I think you may be wondering what is the zero of the polynomial.

But this zero of the polynomial states that when ever this value is substituted in the expression then the result will be zero(0).

So, let's coming to the question.

( i ) p (y) = 2y^2 + 5y - 3

Let assume that the zero of the polynomial be as -1/2

so,

$p( \frac{ - 1}{2} ) = 2( \frac{ -1}{2} {)}^{2} + 5( \frac{ -1}{2} ) - 3 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2( \frac{1}{4} ) - \frac{5}{2} - 3 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} - \frac{5}{2} - \frac{3}{1} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1 - 5 - 6}{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 10}{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - 5$

Therefore,

Our assumption is wrong

hence, -1/2 is not the zero of the polynomial .

( ii ) p (z) = z^2 + z - 6

Let assume that the zero of the polynomial be as -3

so,

$p( - 3) = ( - 3 {)}^{2} + ( - 3) - 6 \\ \\ \: =9 - 3 - 6 \\ \\ \: = 9 - 9 \\ \\ \: = 0$

Therefore,

Our assumption is correct .

Hence , -3 is the zero of the polynomial.

( iii ) p (x) = 2x^3 - 3x^2 + 5 - 1

Let assume that the zero of the polynomial be as 1.

so,

$p(1) = 2(1 {)}^{3} - 3(1 {)}^{2} + 5(1) - 1 \\ \\ = 2(1) - 3(1) + 5 - 1 \\ \\ = 2 - 3 + 5 - 1 \\ \\ = 7 - 4 \\ \\ = 3$

Therefore,

Our assumption is wrong

Hence, 1 is not zero of the polynomial

( iv ) p (t ) = 2x^3 - 9x^2 + x + 12

Let assume the zero of the polynomial be 3/2

$p( \frac{3}{2} ) = 2( \frac{3}{2} {)}^{3} - 9( \frac{3}{2} {)}^{2} + \frac{3}{2} + 12 \\ \\ = 2( \frac{27}{8} ) - 9( \frac{9}{4} ) + \frac{3}{2} + 12 \\ \\ = \frac{27}{4} - \frac{81}{4} + \frac{3}{2} + 12 \\ \\ = \frac{27 - 81 + 6 + 48}{4} \\ \\ = \frac{81 - 81}{4} \\ \\ = \frac{0}{4} \\ \\ = 0$

Therefore,

Our assumption is correct.

Hence, 3/2 is the zero of the polynomial.