Question

If p(x)=x3-2x2+kx+5 is divided by (x-2),the remainder is 11.Find k.Hence find all zeros of x3+kx2+3x+1.

Answer 1

Please refer the attachment dear mates...

Answer 2

Answer:

The value of k is 3

$\text{The zeroes are x=-1,-1,-1}$

Step-by-step explanation:

Given the polynomial

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

when the above polynomial is divided by (x-2), the remainder is 11

we have to find the value of k.

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

By remainder theorem

$p(2)=11$

$2^3-2(2^2)+2k+5=11$

$8-8+2k+5=11$

$2k=11-5=6$

$k=\frac{6}{2}=3$

Now, we have to find the zeroes of the polynomial

$x^3+3x^2+3x+1$

By cube of sum rule

$a^3+3a^2b+3ab^2+b^3=(a+b)^3$

Put a=x and b=1

$x^3+3x^2+3x+1=(x+1)^3$

$x^3+3x^2+3x+1=0$

$(x+1)^3=0$

Hence, the zeroes are

$x=-1,-1,-1$