Question

find the value of k if 2x-1 is a factor of f(x)=2x^3-kx^2-x+2

Answer 1

Step-by-step explanation:

$\huge\fcolorbox{cyan}{blue}{Answer} </p><p>$

P(x) = 6x² + kx - 2

g(x) = 2x - 1

Put g(x) = 0

2x - 1 = 0

x = 1/2

Therefore, g(x) is a factor of p(x).

Then, p( 1/2) = 0

$</p><p>6 {( \frac{1}{2} )}^{2} + k( \frac{1}{2} ) - 2 = 0$

$6 \times \frac{1}{4} - 2+ \frac{1}{2} k = 0 \\ \\ 6× </p><p>4</p><p>1</p><p> </p><p> −2+ </p><p>2</p><p>1</p><p> </p><p> k=0</p><p> \\ </p><p> \\ </p><p>\frac{3}{2} - 2 + \frac{1}{2} k = 0 </p><p>2</p><p>3</p><p> </p><p> −2+ </p><p>2</p><p>1</p><p> </p><p> k=0</p><p></p><p>$

$\frac{1}{2} </p><p> </p><p> + </p><p> \frac{1}{2} </p><p> </p><p> k=0$

$\frac{1}{2} \: k \: = \frac{1}{2}$

$k \: = \frac{1}{2} \times 2$

$k \: = \: 1$

Answer 2

P(x) = 6x² + kx - 2

g(x) = 2x - 1

Put g(x) = 0

2x - 1 = 0

x = 1/2

Therefore, g(x) is a factor of p(x).

Then, p( 1/2) = 0

< /p > < p > 6 {( \frac{1}{2} )}^{2} + k( \frac{1}{2} ) - 2 = 0</p><p>6(

2

1

)

2

+k(

2

1

)−2=0

\begin{gathered}6 \times \frac{1}{4} - 2+ \frac{1}{2} k = 0 \\ \\ 6× < /p > < p > 4 < /p > < p > 1 < /p > < p > < /p > < p > −2+ < /p > < p > 2 < /p > < p > 1 < /p > < p > < /p > < p > k=0 < /p > < p > \\ < /p > < p > \\ < /p > < p > \frac{3}{2} - 2 + \frac{1}{2} k = 0 < /p > < p > 2 < /p > < p > 3 < /p > < p > < /p > < p > −2+ < /p > < p > 2 < /p > < p > 1 < /p > < p > < /p > < p > k=0 < /p > < p > < /p > < p > \end{gathered}

6×

4

1

−2+

2

1

k=0

6×</p><p>4</p><p>1</p><p></p><p>−2+</p><p>2</p><p>1</p><p></p><p>k=0</p><p>

</p><p>

</p><p>

2

3

−2+

2

1

k=0</p><p>2</p><p>3</p><p></p><p>−2+</p><p>2</p><p>1</p><p></p><p>k=0</p><p></p><p>

\frac{1}{2} < /p > < p > < /p > < p > + < /p > < p > \frac{1}{2} < /p > < p > < /p > < p > k=0

2

1