Hi,
$f(1)=2*1^5-9*1^4+12*1^3-12*1^2+10*1-3\\
=2-9+12-12+10-3\\
=0\\

f(x)=(x-1)(2x^4-7x^3+5x^2-7x+3)\\


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

g(3)=2*3^4-7*3^3+5*3^2-7*3+3\\
=162-189+45-21+3\\
=0\\

g(x)=(x-3)(2x^3-x^2+2x-1)\\

h(x)=2x^3-x^2+2x-1\\

h( \dfrac{1}{2} )=2* \dfrac{1}{8} - \dfrac{1} {4}+1-1\\
=0\\

h(x)=(2x-1)(x^2+1)\\


f(x)=(x-1)(x-3)(2x-1)(x+i)(x-i)\\


$
Roots are 1,3,0.5,i,-i.

8 0

2 years ago

In a certain county, the number of Charter Schools is eleven less than twice the number of Alternative Schools. There are 35 Cha

vitfil [10]

Id have to say 2a-11=35

5 0

2 years ago

What are the zeros of this graph

kotegsom [21]

The zeroes ( where the graph cuts the x axis) ar (-2,0) AND (2,0)

tHE FACTOrIAL FORM IS (x - 2)(x + 2)

Its B

3 0

3 years ago

Can some one answer quickly please

denis-greek [22]

Answer:

the answer is A -1/3 x

Step-by-step explanation:

hope this helps :)

3 0

2 years ago