Question

The Cubic polynomial f(x) is such that the coefficient of x ^ 3 is -1 and the zeros of f(x) are 1, 2 and k. If f (x) has a remainder of 8 when divided by x - 3 then find (1) the value of k (2) the remainder when f(x ) is divided by x + 3

Answer 1

Answer:

1)k = 7

2)200

Step-by-step explanation:

Given the zeros of f(x) are 1,2 and k.

Any cubic polynomial with 3 roots p,q and r  and leading coefficient a will be in the form

a(x-p)(x-q)(x-r) = 0

Given f(x) has leading coefficient -1 so a = -1

Given the zeros p =1 , q =2 and r = k

Hence f(x) will be of the form

f(x) = -(x-1)(x-2)(x-k)

Now, given that f(x) has remainder of 8 when divided by x-3

We know from Remainder's Theorem , when f(x) is divided by (x-a), then the remainder will be equal to f(a).

Hence f(3) = 8

But f(3) = -(2)(1)(3-k) = 8

=>3-k = -4

=>k = 7-----Ans

2)

So, f(x) = -(x-1)(x-2)(x-7)

Now, if f(x) is divided by (x+3), remainder will be equal to

=f(-3)

=-(-4)(-5)(-10)

=200.-----Ans

Answer 2

Answer: k = -1
f(-3) = -40

Solution:

Standard cubic polynomial is given as

$a {x}^{3} + b {x}^{2} + cx + d \\ \\ given \: \: a = - 1 \\ \\ - 1 {x}^{3} + b {x}^{2} + cx + d$
if
$\alpha \beta \gamma$
are the roots of cubic polynomial

$\alpha = 1 \\ \\ \beta = 2 \\ \\ \gamma = k$
$\alpha + \beta + \gamma = \frac{ - b}{a} \\ 3 + k = \frac{ - b}{a} \\ \\ 3 + k = b \\ \\ 2 + k + 2k = \frac{c}{a} \\ \\ 2 + 3k = \frac{c}{a} \\ \\ 2 + 3k = - 1 \\ \\ 2k = \frac{ - d}{a} \\ \\ 2k = d$
cubic polynomial is

${x}^{3} - (3 + k) {x}^{2} + (2 + 3k)x - 2k$
1) Now the polynomial had remainder 8 if divided by x-3
$x - 3){x}^{3} - (3 + k) {x}^{2} + (2 + 3k)x - 2k( {x}^{2} - kx + 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: {x}^{3} - 3 {x}^{2} \\ - - - - - - - - \\ \: \: \: \: \: \: \: \: - k {x}^{2} + (2 + 3k)x \\ \: \: \: \: \: \: \: \: \: - k {x}^{2} + 3kx \\ \: \: \: - - - - - - - \\ \: \: \: \: \: \: \: \: \: 2x - 2k \\ \: \: \: \: \: \: \: \: \: 2x - 6 \\ - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: - 2k + 6$

So

$- 2k + 6 = 8 \\ \\ - 2k = 2 \\ \\ k = - 1$

So the polynomial is

${x}^{3} - (3 - 1) {x}^{2} + (2 - 3)x + 2 \\ \\ {x}^{3} - 2 {x}^{2} - x + 2$
2)
$x + 3){x}^{3} - 2 {x}^{2} - x + 2( {x}^{2} - 5x + 14\\ \: \: \: \: \: \: \: \: \: \: \: \: {x}^{3} + 3 {x}^{2} \\ - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: - 5 {x}^{2} - x + 14\\ \: \: \: \: \: \: \: \: \: \: \: - 5 {x}^{2} - 15x \\ - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: 14x + 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: 14x + 42 \\ - - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: - 40$