To find the value of "a" such that the two polynomials leave the same remainder when divided by x-a, we can set up the equation:
x^3 - x^2 + 4x = (x-a)(x^2 + bx + c) + d ........(1)
x^3 + 6x - 10 = (x-a)(x^2 + bx + c) + d ........(2)
where b, c, and d are constants to be determined.
Since the remainder is the same for both polynomials, we can equate the right sides of equations (1) and (2):
(x-a)(x^2 + bx + c) + d = (x-a)(x^2 + bx + c) + d
Expanding both sides, we get:
x^3 - x^2 + 4x = x^3 + 6x - 10
Rearranging the terms, we have:
x^3 - x^3 - x^2 + 4x - 6x + 10 = 0
- x^2 - 2x + 10 = 0
Simplifying further:
x^2 + 2x - 10 = 0
This quadratic equation can be factored as:
(x + 5)(x - 2) = 0
Therefore, the two values for "a" are -5 and 2.
The polynomial x^3-x^2+4x and x^3 +6x-10 leaves the same remainder when divided byx-a .find the value of a
5 answers
Plot the graph of y=3-x and y=3tanx on the same plane
To plot the graph of y = 3 - x and y = 3tan(x) on the same plane, we can follow these steps:
Step 1: Determine the domain and range for each function:
- The domain for y = 3 - x is all real numbers.
- The range for y = 3 - x is also all real numbers.
- The domain for y = 3tan(x) is all values of x except where tan(x) is undefined. Since tan(x) is undefined at odd multiples of π/2, the domain is x ≠ (2n + 1)π/2, where n is an integer.
- The range for y = 3tan(x) is all real numbers.
Step 2: Find some points to plot for each function:
- For y = 3 - x:
- When x = 0, y = 3 - 0 = 3. So one point is (0, 3).
- When x = 1, y = 3 - 1 = 2. So another point is (1, 2).
- For y = 3tan(x):
- We will find points for one period, from x = -π/2 to x = π/2.
- When x = -π/2, y = 3tan(-π/2) = 3∙(-∞) = -∞.
- When x = -π/4, y = 3tan(-π/4) = 3∙(-1) = -3.
- When x = 0, y = 3tan(0) = 3∙0 = 0.
- When x = π/4, y = 3tan(π/4) = 3∙1 = 3.
- When x = π/2, y = 3tan(π/2) = 3∙(∞) = ∞.
Step 3: Plot the points and graph each function on the same plane.
Here is the graph of y = 3 - x (in blue) and y = 3tan(x) (in red):
*[illustration graph]
Step 1: Determine the domain and range for each function:
- The domain for y = 3 - x is all real numbers.
- The range for y = 3 - x is also all real numbers.
- The domain for y = 3tan(x) is all values of x except where tan(x) is undefined. Since tan(x) is undefined at odd multiples of π/2, the domain is x ≠ (2n + 1)π/2, where n is an integer.
- The range for y = 3tan(x) is all real numbers.
Step 2: Find some points to plot for each function:
- For y = 3 - x:
- When x = 0, y = 3 - 0 = 3. So one point is (0, 3).
- When x = 1, y = 3 - 1 = 2. So another point is (1, 2).
- For y = 3tan(x):
- We will find points for one period, from x = -π/2 to x = π/2.
- When x = -π/2, y = 3tan(-π/2) = 3∙(-∞) = -∞.
- When x = -π/4, y = 3tan(-π/4) = 3∙(-1) = -3.
- When x = 0, y = 3tan(0) = 3∙0 = 0.
- When x = π/4, y = 3tan(π/4) = 3∙1 = 3.
- When x = π/2, y = 3tan(π/2) = 3∙(∞) = ∞.
Step 3: Plot the points and graph each function on the same plane.
Here is the graph of y = 3 - x (in blue) and y = 3tan(x) (in red):
*[illustration graph]
#1. Your remainders when a = -5 are not the same.
Using the Remainder Theorem, the remainders when divided by (x-a) are f(a), so we have
a^3-a^2+4a = a^3+6a-10
a^2+2a-10 = 0
a = -1±√11
I suspect a typo, since the answer is so unexpected.
Using the Remainder Theorem, the remainders when divided by (x-a) are f(a), so we have
a^3-a^2+4a = a^3+6a-10
a^2+2a-10 = 0
a = -1±√11
I suspect a typo, since the answer is so unexpected.
I apologize for the mistake in my previous response. You are correct, the value of "a" should be -1 ± √11. Thank you for pointing that out.
4 answers