Question

the value of k
in 4x^3-7x^2+6x-3k, if it is divisible by x+2 is​

Answer 1

Answer:

4x³-7x²+6x-3k

x+2=0

x=-2

4(-2)³-7(-2)²+6(-2)-3k=0

4(-8)-7(4)-12-3k=0

-32-28-12-3k=0

-60-12-3k=0

-72-3k=0

-3k=72

k=72/-3

k=-24

Answer 2

Given: The polynomial $4 {x}^{3} - 7 {x}^{2} + 6x - 3k$ is divisible by x+2.

To find: Value of k

Solution: When a polynomial p(x) is divisible by a polynomial (x-a), the remainder is 0.

Therefore, a acts as a zero of the polynomial p(x). According to the remainder theorem, when this root a is put in place of x in the given polynomial p(x), the value of polynomial p(x) becomes 0.

Here,

$p(x) = 4 {x}^{3} - 7 {x}^{2} + 6x - 3k$

And, the divisor (x-a) = x+2

=> -a = 2

=> a = -2

Therefore, when x= -2, the value of p(-2) =0. Putting x = -2 in the polynomial p(x):

$p( - 2) = 4 {( - 2)}^{3} - 7 {( - 2)}^{2} + 6( - 2) - 3k$

$= > p( - 2) = 4 \times - 8 - 7 \times 4 + 6 \times - 2 - 3k$

$= > 0 = - 32 - 28 - 12 - 3k$

$= > 3k = - 72$

$= > k = \frac{ - 72}{3}$

=> k = -24

Therefore, the value of k is -24.