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
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.
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]
#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.
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.
To find the value of "a" such that both polynomials leave the same remainder when divided by x-a, we can set up an equation.
Let's divide each polynomial by x-a and set their remainders equal to each other:
For the polynomial x^3 - x^2 + 4x:
x^3 - x^2 + 4x = Q(x) * (x - a) + R1, where Q(x) is the quotient and R1 is the remainder.
For the polynomial x^3 + 6x - 10:
x^3 + 6x - 10 = Q(x) * (x - a) + R2, where Q(x) is the quotient and R2 is the remainder.
Since both polynomials leave the same remainder, we can set R1 equal to R2:
R1 = R2
Now, let's substitute the expressions for R1 and R2:
x^3 - x^2 + 4x = Q(x) * (x - a) + x^3 + 6x - 10
Rearranging the equation, we have:
x^3 - x^2 + 4x - x^3 - 6x + 10 = Q(x) * (x - a)
Simplifying further:
- x^2 - 2x + 10 = Q(x) * (x - a)
Since this equation should hold for all values of x, we can equate the coefficients on both sides. Specifically, we can equate the coefficients of each power of x.
For the constant terms:
10 = Q(x) * (-a)
Since Q(x) can be any polynomial, it follows that Q(x) must be 1 when multiplied by -a.
Therefore, a = -10.
Hence, the value of "a" is -10.
To find the value of 'a' such that both polynomials leave the same remainder when divided by x-a, we need to set up and solve the equation.
First, let's perform the division of each polynomial by x-a and express the remainders:
For x^3 - x^2 + 4x divided by x-a:
x^3 - x^2 + 4x = (x-a)(x^2 + (a-1)x + (a^2 - a))
The remainder is (a^2 - a).
For x^3 + 6x - 10 divided by x-a:
x^3 + 6x - 10 = (x-a)(x^2 + ax + (a^2 + 2a + 10))
The remainder is (a^2 + 2a + 10).
Since both polynomials leave the same remainder, we can equate the remainders and solve for 'a':
(a^2 - a) = (a^2 + 2a + 10)
This equation simplifies to:
-a = 2a + 10
Combining like terms:
3a = -10
Finally, solving for 'a':
a = -10/3
Therefore, the value of 'a' is -10/3.