Question

1.P(y)= 4+3y-y2+5y3 at y=0,1,2,-1.
• 2.If f(t)= 4t2-3t +6 at t = 0,4,-5.
• 3.If p(x)= x2-3x2+2x at x= 0,1,2.
What do you conclude?
4. Verify that 1 and 2 are the zeroes
of polynomial, p(x) = x2-3x+2.
5. Verify that 2 and -3 are the
zeroes of the polynomial, 9(x)=
x2+x-6.​

Answer 1

Answer:

1) P(0)=4+3×0-0^2+5×0^3

= 4

P(1)=4+3×1-1^2+5×1^3

=4+3-1+5

=12-1

=11

P(2)=4+3×2-2^2+5×2^3

=4+6-4+5×6

=6+30=36

P(-1)=4+3×(-1)-(-1)^2+5×(-1)^3

=4-3-1-5

=4-9=-5

Q4)P(x)=x2-3x+2

if 1 and 2 are the zeroes of p(x), P(1)=0,P(2)=0

Now,P(1)=1^2-3×1+2

=1-3+2

=3-3 =0

P(2)=2^2-3×2+2

=4-6+2

=6-6=0

since P(1)=0 and P(2)=0,so 1 and 2 are the zeroes of given polinomial P(x)