Math, asked by XxLUCYxX, 1 month ago

Find the value of ‘k’ if 4x 3 – 2x 2 + kx + 5 leaves remainder -10 when divided by 2x + 1.


please help​

Answers

Answered by diyaadhana50
1

Answer:

let,

f(x) = 4x^{3}-2 x^{2} + kx +5f(x)=4x3−2x2+kx+5

it is divided by 2x+1=0

⇒ value of x=-1/2

(u can find it out by:

 2x+1=0 ∴ x=-1/2

)

remainder=-10

f(- \frac{1}{2} )=4( - \frac{1}{2}^{3} )-2( - \frac{1}{2})^{2} +k(- \frac{1}{2})+5=-10f(−21)=4(−213)−2(−21)2+k(−21)+5=−10

solving the above equation ans is : k=28

Answered by kamalhajare543
33

Answer:

Answer:

Answer:

\sf \: The \: value \: of \: k \: is \: \bold{ 28.}

Step-by-step explanation:

We have the following theorem :

Remainder Theorem : When a polynomial p(x) is divided by a linear factor (x-a), then the remainder is p(a).

The given polynomial

\sf \: p(x)=4x^3-2x^2+kx+5p(x)=4x +kx+5 \\ \\ \sf \: leaves \: remainder \: -10, \: when \: divided \: by \: 2x+1.

Now,

\sf \: 2x+1=0\\\\\Rightarrow \sf \: x=-\dfrac{1}{2}.

Therefore, we must have

\begin{gathered}\begin{gathered} \sf \: p(\frac{1}{2})=-10\\\\ \sf \implies \sf \: 4\times(-\frac{1}{2})^3-2\times(-\frac{1}{2})^2+k\times(-\frac{1}{2})+5=-10\\\\\\\sf \implies-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{k}{2}+5=-10\\ \\ \\ \sf \implies \sf \: -\dfrac{k}{2}+4=-10\\ \\ \sf \implies \sf-\dfrac{k}{2}=--14\\ \\ \sf \implies \sf \boxed{ \red{ \sf k=28}}.\end{gathered}\end{gathered}

Thus, the required value of k is 28

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