Question

fill in the blank a.11/16_14/15 b.8/15_95/14 c.12/75_32/200​

Answer 1

Answer:

Since, it is given that

3

and −

3

are the zeroes of the polynomial f(x)=x

3

−4x

2

−3x+12, therefore, (x−

3

) and (x+

3

) are also the zeroes of the given polynomial. Now, consider the product of zeroes as follows:

(x−

3

)(x+

3

)

=(x)

2

−(

3

)

2

(∵a

2

−b

2

=(a+b)(a−b))

=x

2

−3

We now divide x

3

−4x

2

−3x+12 by (x

2

−3) as shown in the above image:

From the division, we observe that the quotient is x−4 and the remainder is 0.

Since x−4 is the quotient,

Hence, the third zero of f(x)=x

3

−4x

2

−3x+12 is 4.

solution

Step-by-step explanation:

Since, it is given that

3

and −

3

are the zeroes of the polynomial f(x)=x

3

−4x

2

−3x+12, therefore, (x−

3

) and (x+

3

) are also the zeroes of the given polynomial. Now, consider the product of zeroes as follows:

(x−

3

)(x+

3

)

=(x)

2

−(

3

)

2

(∵a

2

−b

2

=(a+b)(a−b))

=x

2

−3

We now divide x

3

−4x

2

−3x+12 by (x

2

−3) as shown in the above image:

From the division, we observe that the quotient is x−4 and the remainder is 0.

Since x−4 is the quotient,

Hence, the third zero of f(x)=x

3

−4x

2

−3x+12 is 4.

solution

Answer 2

Answer:

$(a) \: \frac{11}{16} < \frac{14}{15} \\ (b) \: \frac{8}{15} < \frac{95}{14} \\ (c) \: \frac{12}{75} - \frac{32}{200} \\ = > \frac{4}{25} = \frac{4}{25} \\ \\ \red{\mathfrak{\underline{\large{Hope \: It \: Helps \: You }}}} \\ \green{\mathfrak{\underline{\large{Mark \: Me \: Brainliest}}}}$