<h2>
For a = 1 and b = 10 x+1 and x+2 factors of x³-ax²-bx-8 = 0</h2><h2>
Other factor is (x-4)</h2>
Step-by-step explanation:
We have
x³-ax²-bx-8 = 0
Its factors are x+1 and x+2
That is x = -1 and x = -2 are its roots
Substituting x = -1
(-1)³-a(-1)²-b(-1)-8 = 0
-1 - a + b - 8 = 0
b - a = 9 ---------------------eqn 1
Substituting x = -2
(-2)³-a(-2)²-b(-2)-8 = 0
-8 - 4a + 2b - 8 = 0
2b - 4a = 16 ---------------------eqn 2
eqn 1 x -2
-2b + 2a = -18 ---------------------eqn 3
eqn 2 + eqn 3
-2a = -2
a = 1
Substituting in eqn 1
b - 1 = 9
b = 10
For a = 1 and b = 10, x+1 and x+2 factors of x³-ax²-bx-8 = 0
The equation is x³-x²-10x-8 = 0
Dividing with x + 1 we will get
x³-x²-10x-8 = (x+1)(x²-2x-8)
Dividing (x²-2x-8) with x + 2 we will get
x²-2x-8 = (x+2)(x-4)
So we have
x³-x²-10x-8 = (x+1)(x+2)(x-4)
Other factor is (x-4)
Answer:
1615 / 48
Step-by-step explanation:
So the mixed numbers as improper fractions is:
6 1/3 = 19 / 3
4 1/4 = 17 / 4
1 1/4 = 5 / 4
19 / 3 * 17 / 4 * 5 / 4 = 1615 / 48
It's kinda messy so it might be wrong.
Your awnser is 120 if I’m wrong please tell me and I’ll help you again
Let's do this by Briot-Ruffini
First: Find the monomial root
x - 2 = 0
x = 2
Second: Allign this root with all the other coeficients from equation
Equation = -3x³ - 2x² - x - 2
Coeficients = -3, -2, -1, -2
2 | -3 -2 -1 -2
Copy the first coeficient
2 | -3 -2 -1 -2
-3
Multiply him by the root and sum with the next coeficient
2.(-3) = -6
-6 + (-2) = -8
2 | -3 -2 -1 -2
-3 -8
Do the same
2.(-8) = -16
-16 + (-1) = -17
2 | -3 -2 -1 -2
-3 -8 -17
The same,
2.(-17) = -34
-34 + (-2) = -36
2 | -3 -2 -1 -2
-3 -8 -17 -36
Now you just need to put the "x" after all these numbers with one exponent less, see
2 | -3x³ - 2x² - 1x - 2
-3x² - 8x - 17 -36
You may be asking what exponent -36 should be, and I say:
None or the monomial. He's like the rest of this division, so you can say:
(-3x³ - 2x² - x - 2)/(x - 2) = -3x² - 8x - 17 with rest -36 or you can say:
(-3x³ - 2x² - x - 2)/(x - 2) = -3x² - 8x - 17 - 36/(x - 2)
Just divide the rest by the monomial.
