Question

if 0 and 2 are the zeroes of the polynomial p[x]=x3-5x2-ax+b,then find the values of a and b.

Answer 1

Answer:

p(0)=0(3)-5×2-a(0)+b=0

0-10-0+b=0

b=10

p(2)=2(3)-5×2-a(2)+10

=6-10-2a+10

=6-2a=0

=-2a=-6

=a=-6÷-2

=a=3

Answer 2

Answer:

If 0 and 2 are the zeroes of the polynomial p[x]=x3-5x2-ax+b, then the values of a and b is -6 and 0 respectively.

Step-by-step explanation:

If 0 and 2 are the zeroes of the polynomial $p[x]=x^3-5x^2-ax+b$

If x  = 0

Then

$p[0]=(0)^3-5(0)^2-a(0)+b\\0 - 0- 0 + b = 0\\b= 0$

Now x = 2

Then

$p[2]=(2)^3-5(2)^2-a(2)+(0)\\8 - 20- 2a= 0\\\\-12 - 2a = 0\\-2a = 12\\a = 12/-2 \\a= -6$