Question

When f(x) = x³ + 3x² – kx + 4 is divided by (x – 2), the remainder is k.i) Find the value of k.ii) Check (x – 1) is a factor of f(x).

Answer 1

➡️Answer:

k= 12

(x-1) is not a factor of f(x)

➡️Solution:

i) Find the value of k:

$(x - 2) \: is \: a \: factor \: of \: {x}^{3} + 3 {x}^{2} - kx + 4$
than it divides completely,i. e. Remainder will be zero

$x - 2) {x}^{3} + 3 {x}^{2} - kx + 4( {x}^{2} + 5x + 10 - k \\ \: \: \: \: \: \: \: \: \: \: \: \: {x}^{3} - 2 {x}^{2} \\ - - - - - - - - change \: sign\\ \: \: \: \: \: \: \: \: \: 5 {x}^{2} - kx \\ \: \: \: \: \: \: \: \: \: \: \: 5 {x}^{2} - 10x \\ - - - - - - - - - - \\ \: \: \: \: \: \: \: \: (10 - k)x + 4 \\ \: \: \: \: \: \: \: \: \: (10 - k)x - 20 + 2k \\ \: \: \: \: \: \: \: \: - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 24 - 2k$

So

$24 - 2k = 0 \\ \\ 2k = 24 \\ \\ k = 12$

Now the polynomial is

$f(x) = {x}^{3} + 3 {x}^{2} - 12x + 4$

2) If (x-1) is a factor of f(x) ,so on putting x=1, it becomes zero

$f(1) = {(1)}^{3} + 3( {1)}^{2} - 12(1) + 4 \\ \\ = 1 + 3 - 12 + 4 \\ \\ f(1)= - 4$
No, (x-1) is not a factor of f(x)

Hope it helps you.

Answer 2

Answer:

Step-by-step explanation:

Concept:

Factor theorem:

(x-a) is a factor of f(x) if and only if f(a)=0

Remainder theorem;

when f(x) is divided by (x-a), the remainder is f(a)

1. f(x) = x³ + 3x² – kx + 4

The remainder when f(x) is divided by (x-2) is k

By remainder theorem,

f(2)=k

(2)³ + 3(2)² – k(2) + 4 = k

8+12-2k +4 = k

24 - 2k= k

24 = 3k

k=8

2.. f(x) = x³ + 3x² – kx + 4

put x=1,

f(1) = (1)³ + 3(1)² – 8(1) + 4

f(1) = 1+3-8+4 =0

since f(1)=0, by factor theorem

(x-1) is a factor of f(x)