Question

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Answer 1

Answer:

i tried to solve the maximum of the questions

Step-by-step explanation:

8.x2-6x+8

p(2)

put x=2

p(2)=(2*2)-6*-2+8

=4-12+8

=0

p(-2)

put x=-2

now

(-2*-2)-6*-2+8

=4+12+8

=24

9)

now

3x^2-2x)6x^4+5x^3+3x-5(2x^2+3x+2

              6x^4-4x^3

---------------------------------------

9x^3+3x-5

9x^3-6x^2

----------------------------

6x^2+3x-5

6x^2-4x

-----------------

7x-5

now divisoin rule

6x^4+5x^3+3x-5=(3x^2-2x)(2x^2+3x+2)+(7x-5)

10)

given

f(x)=x^3-9x^2+26x-24

now f(2)=2^3-9*(2^2)+26*2-24

=8-36+52-24

=60-60=0

f(3)=(3^3)-(9*(3^2))+26*3-24

=27-81+78-24

=0

f(4)=(4^3)-(9*(4^2))+26*4-24

=64-144+104-24

=0

so 2,3,4, are zeros of f(x)

here the relation between zeros and its coefficients are the second coefficient is the square of one of its zeros and the last constant is the product of all the zeros