Answer:
Below.
Step-by-step explanation:
Substitute x = 3 into each polynomial and see if the value of the polynomial is zero. If it is zero then x-3 is a factor.
Try number 1:
a(x) = x^3 - 2x^2 - 4x + 3
a(3) = 3^3 - 2(3)^2 - 4(3) + 3
= 27 - 2(9) - 12 + 3
= 27 - 18 - 12 + 3
= 9 - 18 - 9
= 27 - 27
= 0
So a(x) has (x-3) is a factor of a(x)
The other 3 polynomials are solved in the same way.
Rewrite g(x) as x-1
------
4
and then substitute this result for x in f(x) = x^2 - 3x + 3:
f(g(x)) = (x-1)^2 / 4^2 - 3(x-1)/4 + 3.
At this point we can substitute the value 5 for x:
f(g(5)) = (5-1)^2 / 4^2 - 3(5-1)/4 + 3
= 16/16 - 3(4/4) + 3 = 1 - 3 + 3 = 1
Therefore, f(g(5)) = 1.
I’m not too sure if I did it correct, but the answer I got was -6x^2+36. I just assumed you’d solve it as usual and then multiply (-3) after distributing-2 to x^2-6. Hope this helps.
Answer:


Step-by-step explanation:
Given system of equations:

To solve by the method of substitution, substitute the first equation into the second equation and rearrange so that the equation equals zero:

Factor the quadratic:

Apply the <u>zero-product property</u> and solve for x:


Substitute the found values of x into the <u>second equation</u> and solve for y:


Therefore, the solutions are:


Wow I don't get it either but my guess would be 9,6...sorry....