See explanation below.
Explanation:
The 'difference between roots and factors of an equation' is not a straightforward question. Let's define both to establish the link between the two..
Assume we have some function of a single variable
x
;
we'll call this
f
(
x
)
Then we can form an equation:
f
(
x
)
=
0
Then the "roots" of this equation are all the values of
x
that satisfy that equation. Remember that these values may be real and/or imaginary.
Now, up to this point we have not assumed anything about
f
x
)
. To consider factors, we now need to assume that
f
(
x
)
=
g
(
x
)
⋅
h
(
x
)
.
That is that
f
(
x
)
factorises into some functions
g
(
x
)
×
h
(
x
)
If we recall our equation:
f
(
x
)
=
0
Then we can now say that either
g
(
x
)
=
0
or
h
(
x
)
=
0
.. and thus show the link between the roots and factors of an equation.
[NB: A simple example of these general principles would be where
f
(
x
)
is a quadratic function that factorises into two linear factors.
Answer:
I believe it is B
Step-by-step explanation:
3 out of 4 options to land on are less than 5
Answer:
Step-by-step explanation:
Step 1: Identify the GCF of the polynomial.
Step 2: Divide the GCF out of every term of the polynomial. ...
Step 1: Identify the GCF of the polynomial. ...
Step 2: Divide the GCF out of every term of the polynomial.
Step 1: Identify the GCF of the polynomial. ...
Step 2: Divide the GCF out of every term of the polynomial .
The inverse of this function is not a function and this function isn’t inversable because it’s not one to one
Answer:
d. (x+2)/(-x²-5)
Step-by-step explanation
ƒ(x) = x + 2/(2x²)
The function is undefined when x = 0.
b. ƒ(x) = (2x + 4)/(3x + 3)
The function is undefined when 3x + 3 = 0, i.e., when x = -1.
c. ƒ(x) = (6x - 5)/(x² - 7)
The function is undefined when x² - 7 = 0, i.e., when x = √7.
d. ƒ(x) = (x+2)/(-x²-5) = -(x+2)/(x² + 5)
The function would be undefined if x² + 5 = 0, i.e., if x² = -5. However, the square of a real number cannot be negative.
This function has no excluded values.
