Question

For what value of k, -7 is the zero of the polynomial 2x^2+11x+(6k-3). Also find the other zero of the polynomial​

Answer 1

for -7 to be root
2*49-77+6k-3=0
⇒98-77-3+6k=0
⇒18+6k=0
⇒6k=-18
⇒k=-3

Answer 2

Answer:

The value of k is -3 and the other zero is $\frac{3}{2}$

Step-by-step explanation:

Given : -7 is the zero of the polynomial$2x^2+11x+6k-3$.

To Find : Value of k and Also find the other zero of the polynomial​

Solution:

Since - 7 is the zero of the polynomial

So, Substitute x= -7

$2x^2+11x+6k-3$

$2(-7)^2+11(-7)+6k-3=0$

$18+6k=0$

$-3=k$.

So, Polynomial=$2x^2+11x+6(3)-3$.

                       =$2x^2+11x-21$.

Since -7 is the zero of the polynomial

So, Divisor = x+7

Dividend = $2x^2+11x-21$

We know that $Dividend = (Divisor \times quotient)+Remainder$

$2x^2+11x-21 = (x+7 \times 2x-3)+0$

Quotient = 2x-3

So, 2x-3=0

$x=\frac{3}{2}$

So, The value of k is -3 and the other zero is $\frac{3}{2}$