<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)
62 inchezzzzzzz so 60-4 = 56 --- 56 +6 = 62
Answer:
14
Step-by-step explanation:
The computation of the mean number of coins for the girls is shown below:
Since there are 24 boys and 16 girls so the total children is 40
So the number of coins for all children is
= 40 × 11
= 440
And, the number of coins for all boys is
= 9 × 24
= 216
Now the mean is
= (440 - 216) ÷ 16 girls
= 14
Answer:
3/4
Step-by-step explanation:
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