Question

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

Answer 1

$\textbf{\huge{ANSWER:}}$

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)

Answer 2

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.