Part 1: Answer:
(x+1)(x+1)(x-6) = x^3 - 4x^2 - 11x - 6
Step-by-step explanation:
To make r a root, include (x-r) as a factor. (-1+1)(-1+1)(-1-6) is zero even though (-1-6) isn't.
(6+1)(6+1)(6-6) is zero.
Part 2 Answer:
Standard form: y = -x^4 + 12
Degree 4
left end goes down, right end goes down.
Step by step: apply the definitions of standard form, polynomial degree, and "end behavior". In other words, read the textbook.
Part 3: Answer: x = 3, x = 8
Step by step:
x^2-11x = -24
x^2-11x+24 = 0
(x-3)(x-8) = 0
x = 3 or x = 8
Part 4a Answer:
quotient 2x^2 + x - 3
remainder 1
Step by step:
2x^2 + x - 3
___________________
x-4 ) 2x^3 - 7x^2 - 7x + 13
2x^3 - 8x^2
__________
0 + x^2 - 7x + 13
x^2 - 4x
____________
0 - 3x + 13
- 3x + 12
______
1
Part 4b answer:
quotient 2x^2 - 6x + 2
remainder -20
Step by step: you have to know exactly what you are doing. Refer to textbook or Wikipedia.
dividend 2x^3 +14x^2 - 58x
divisor x+10
leading coefficient of divisor must be 1
write coefficients of dividend at top
write coefficients of dividend at left
| 2 14 -58 0
-10 | -20 60 -20
___________
| 2 -6 2 -20
Coefficients of quotient are 2 -6 2
Remainder is -20
quotient = 2x^2 - 6x + 2
Answer:
See below.
Step-by-step explanation:
4(18-3k)=9(k+1)
72-12k=9k+9
+12k +12k
72=21k+9
-9 -9
63=21k
/21 /21
3=k
-hope it helps
Answer:
Yes , it is separable
Step-by-step explanation:
We are given that DE
We have to find the the given DE is separable or not.
When the DE is separable then it can be written as
Given DE is in Separable form.Therefore, it is separable.
Im assuming its x^2+4x+3.....for x squared right.
First get the product of two numbers that equal to 3 (+3 * +1)
But also add up to 4 (+3 + <span>+1</span>)
so you apply them to your x^2
(x+3)(x+1)
