Question

If one zero of the polynomial 3x2 - 10x +
3 is 1/3, the other zeroes is
A.
-3
B.
3
C.
-1/3​

Answer 1

Answer:

B)

Step-by-step explanation:

Given polynomial : 3x^2- 10 x +33x

2

−10x+3

If one zero is \dfrac{1}{3}

3

1

To find : The other zero of the polynomial

As we know if ax² + bx +c is the standard form of quadratic polynomial

Then comparing we get

a = 3 , b = -10 , c= 3

Now the sum of zeroes is given by

\begin{gathered}\alpha + \beta = \dfrac{-b}{a} \\\\\Rightarrow \dfrac{1}{3} + \beta= \dfrac{-(-10)}{3} \\\\\Rightarrow \dfrac{1 + 3 \beta}{3} = \dfrac{10}{3} \\\\\Rightarrow 1+ 3\beta = 10 \\\\\Rightarrow 3\beta= 9 \\\\\Rightarrow \beta= 3\end{gathered}

α+β=

a

−b

⇒

3

1

+β=

3

−(−10)

⇒

3

1+3β

=

3

10

⇒1+3β=10

⇒3β=9

⇒β=3

Answer 2

Step-by-step explanation:

p(x) = 0

p( 1/3 ) = 3x² - 10x + 3

= 3 ( 1/3 )² + 10 (1/3) + 3

= 3 ( 1/9) + 10 /3 + 3

= 3/9 + 10 /3 + 3

= 1 + 10 + 3

3 3 1

= 1 + 10 + 9

3

= 20

3.