Problem 1 should be B, C, and D. I'm not sure about the rest... Sorry I couldn't be more of a help.
The zeros of m(x) =x^2-4x+3 are 1 and 3
Answer:
4x+x=40
5x+40
x=8
8, 32
Step-by-step explanation:
Answer:
3, 27 and 9,9
Step-by-step explanation:
3*27 = 81
9*9=81
Technically you could say 1 and 81 but those two answers fit the most.
Answer:
3) x = -1, 2, 3
Step-by-step explanation:
2)
x is a root of this function, so when we put x=2 in this function, we'll get f(x)=( 2+1 )*( 2-2 )*( 2-3 ). Because 2-2=0, so the function will equals to zero, too.
3)
By factorizing f (x) = x^3 – 4x^2 + x + 6, we can get ( x+1 )*( x-2 )*( x-3 )=0,
and if one of those three parts is zero, that the whole formula will be zero, now we get : x+1=0 or x-2=0 or x-3=0, so x can be -1, 2, or 3.
//sorry that I only could solve question 2 and 3...