Answer:
x^4 - 14x^2 - 40x - 75.
Step-by-step explanation:
As complex roots exist in conjugate pairs the other zero is -1 - 2i.
So in factor form we have the polynomial function:
(x - 5)(x + 3)(x - (-1 + 2i))(x - (-1 - 2i)
= (x - 5)(x + 3)( x + 1 - 2i)(x +1 + 2i)
The first 2 factors = x^2 - 2x - 15 and
( x + 1 - 2i)(x +1 + 2i) = x^2 + x + 2ix + x + 1 + 2i - 2ix - 2i - 4 i^2
= x^2 + 2x + 1 + 4
= x^2 + 2x + 5.
So in standard form we have:
(x^2 - 2x - 15 )(x^2 + 2x + 5)
= x^4 + 2x^3 + 5x^2 - 2x^3 - 4x^2 - 10x - 15x^2 - 30x - 75
= x^4 - 14x^2 - 40x - 75.
Answer:
Assuming you are talking about segment QS, it's 12 units.
Step-by-step explanation:
10+2=12.
Answer:
m3=56 and m1=75
Step-by-step explanation:
Answer: y=3/2x+7
Step-by-step explanation:
To find the equation of the line that passes through the given point, let's first convert the given equation to slope-intercept form.
3x-2y=2 [subtract both sides by 3x]
-2y=-3x+2 [divide both sides by -2]
y=3/2x-1
Now that we have the equation in slope-intercept form, we know that the slope must stay the same because parallel lines NEVER touch. All we have to do is plug in the given point to find the y-intercept.
-2=3/2(-6)+b [multiply]
-2=-9+b [add both sides by 9]
b=7
Now, we know that the parallel equation is y=3/2x+7.
Yeah you surely can use them by ASA only
