Math, asked by esokiyaaa1883, 1 year ago

If p x is equal to x cube minus 2 x square minus kx + 5 is divided by x minus 2 the remainder is 11 find k hence find all the zeros of x cube + 3 x square + 3 x + 1

Answers

Answered by Anonymous
69
\textbf{\huge{ANSWER:}}

It's \sf{Given} that :-

p (x) = x^{3} - 2x^{2} - kx + 5

And

f (x) = x - 2

Also

p (x) when divided by f (x) leaves remainder 11

---

x - 2 = 0

Solve it further

=》 \boxed{\tt{x = 2}}

Put the value in p (x) :-

=》 2^{3} - 2 (2)^{2} - k(2) + 5 = 11

Solve it further

=》 8 - 8 - 2k + 5 = 11

Some more steps to go

=》 (-2k) = 11 - 5

Last one step

=》 k = ( -3 )

Now, the second part of the question is :-

Find the zeroes of the given polynomial :-

p (x) = x^{3} + 3x^{2} + 3x + 1

Factors of 1 are = (-1) and (+1)

Now, when, x = (-1)

=》 (-1)^{3} + 3(-1)^{2} + 3(-1) + 1

Just simplify it further

=》 (-1) + 3 - 3 + 1

Last one step to go

=》 0

Thus, it's a factor of p (x)

Hence, (x + 1) is a factor of p (x)

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Cubingwitsk: Awesome!
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Anonymous: Nice answer !!
Anonymous: Thanks!
Answered by BrainlyVirat
49
Answer :

\tt{1) p ( x ) = x {}^{ 3 } - 2x {}^{2} - kx + 5}

\tt{f ( x ) = (x - 2)}

Thus,

X - 2 = 0

X = 2

Putting value of x in the polynomial.

\tt{(2 ){}^{3} - 2 \times {2}^{2} - k(2) + 5= 11}

Thus, Simplifying next , we get :

 \tt{8 - 8 - k(2) + 5 = 11}

 \tt{ - 2k + 5 = 11}

Taking coefficients and variables to one side and the other, we get :
 \tt{ - 2k = 11 - 5}

 \tt{ - 2k = 6}

Hence, Dividing both the sides by -2, we get :

 \tt{k = - 3}

Value of k is -3.

Now,

Let's look the next :

\tt{p(x) = x {}^{3} + 3x {}^{2} + 3x + 1}

As we know that Factors of 1 are (-1) & (+1), thus

\tt{( - 1) {}^{3} + 3( - 1) {}^{2} + 3( - 1) + 1}

 \tt{ = - 1 + 3 - 3 + 1}

 \tt {= 2 - 2}

Thus, Final step is :

 \tt{ = 0}

Thus, (x + 1) is the factor of the given polynomial.

Anonymous: Mast answer hai! Perfect!
BrainlyVirat: Thanka! :)
Cubingwitsk: Awesome!
aayush36: are you virat kohli
Anonymous: Great answer bro !!
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