Given that the polynomial f (×)=6-×-×^2 is a factor of the polynomial g (×)=ax^3+5×^2+bx-18=0.find the values of the constants a and g.The remainder when g (x) is divided by x+2
If (6-x-x^2) divides g(x), then that means that (3+x) and (x-2) are both factors of g(x)
g(x)/(x+3) has a remainder of -27a-3b+27
g(x)/(x-2) has a remainder of 8a+2b+2
Since those are both zero, we have
9a+b = 9
4a+b = -1
a=2, b=-9
g(x) = 2x^3+5x^2-9x-18 = (2x+3)(x-2)(x+3)
To find the values of the constants a and b, we can use the fact that f(x) is a factor of g(x).
Step 1: Use polynomial long division to divide g(x) by f(x).
-× - 2
________________
× + 2 ) a×^3 + 5×^2 + b× - 18
-(a×^3 + 2a×^2)
______________
3×^2 + b× - 18
-(3×^2 + 6×)
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-5× + 18
-(-5× + 10)
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8
Step 2: The remainder when g(x) is divided by x + 2 is 8.
Since f(x) is a factor of g(x), the remainder when g(x) is divided by f(x) should be equal to 0. Therefore, the remainder of 8 contradicts f(x) being a factor of g(x).
There must have been an error in the given information or the calculations. Please double-check the polynomials provided.
To find the values of the constants a and b, we can use the fact that f(x) is a factor of g(x) and apply polynomial division.
Given that f(x) = 6 - x - x^2 is a factor of g(x) = ax^3 + 5x^2 + bx - 18, we can perform polynomial division to find the remainder. The remainder when g(x) is divided by x + 2 will be 0 if x + 2 is also a factor of g(x).
To divide g(x) by x + 2, we set up the division as follows:
-2 | a 5 b -18
| -2a -6a+10 -4b-26
+-------------
a-2a 5-6a+10 b-4b-26-18
Since the remainder should be zero, we have:
a - 2a + 5 - 6a + 10 + b - 4b - 26 - 18 = 0
Simplifying this equation, we get:
-7a - 3b - 29 = 0
Now, we have two equations:
a - 2a + 5 - 6a + 10 + b - 4b - 26 - 18 = 0 (Equation 1)
-7a - 3b - 29 = 0 (Equation 2)
We can solve this system of equations to find the values of a and b.
By rearranging Equation 2, we have:
-7a - 3b = 29 (Equation 3)
Now, let's substitute the value of a from Equation 3 into Equation 1:
(-7a - 3b) - 2a + 5 - 6a + 10 + b - 4b - 26 - 18 = 0
Simplifying, we get:
-15a - 6b - 29 = 0
Rearranging this equation, we have:
-15a - 6b = 29 (Equation 4)
Now, we have a system of two equations:
-7a - 3b = 29 (Equation 3)
-15a - 6b = 29 (Equation 4)
We can solve this system of equations to find the values of a and b.
By multiplying Equation 3 by -2 and Equation 4 by 3, we can eliminate b:
14a + 6b = -58 (Equation 5)
-45a - 18b = 87 (Equation 6)
Adding Equation 5 and Equation 6, we get:
-31a = 29
Dividing by -31, we find that a = -29/31.
By substituting the value of a back into Equation 3, we can solve for b:
-7(-29/31) - 3b = 29
203/31 - 3b = 29
-3b = 29 - 203/31
-3b = (29*31 - 203)/31
-3b = 897/31 - 203/31
-3b = 694/31
b = -694/31 * (-1/3)
b = 694/93
Therefore, the values of the constants a and b are:
a = -29/31
b = 694/93