Question

The value of k if (x-1) is a factor of p(x) = 2(x)^2+kx+√2 find the value of k​

Answer 1

Given:-

→A polynomial p(x) = 2x²+kx+√2

→(x-1) is a factor of p(x)

To find:-

→Value of k

Solution:-

By Factor theorem,we know that 'if (x-a) is a factor of p(x), then p(a) = 0, where a is any real number'.

So,now we will use the Factor theorem, to find the value of k.

=>x-1 = 0

=>x = 1

=>p(1) = 0

=>2(1)² + k(1) + √2 = 0

=>2 + k + √2 = 0

=>k = -√2+2

=>k = -(√2+2)

Thus,the value of k is -(√2+2).

Some Extra Information:-

Remainder theorem :- Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. When p(x) is divided by (x-a) then the remainder is p(a).

Answer 2

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find the value of k if (x-1) is a factor of p(x) = 2x²+Kx+√2

$\huge \bold \color{green} \mid{ \underline{ \underline \red{answer}}} \mid$

if (x-1) is a factor of p(x)

so p(1) = 0

now..put the value of x in p(x) = 2x²+Kx+√2

$\implies p(x) = 2 {x}^{2} + kx + \sqrt{2} \\ and \: p(1) = 0 \\ \implies2(1)^{2} + k(1) + \sqrt{2} = 0 \\ \implies2 + k + \sqrt{2} = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \implies \: k = - 2 - \sqrt{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \implies \: k = - (2 + \sqrt{2} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so the value of k is -(2+√2)