Given that the polynomials f (x)=6-x-x^2 is a factor of the polynomials g (x)=ax^3+5x^2+bx=18 find:
The value of the constant a and b
The remainder when g (x)is divided by x+2
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You have
ax^3 + 5x^2 + bx - 18 = (-x^2-x+6)(mx+n)
Clearly,
n=3 and m = -a for the first and last terms to work out
Now, doing the multiplication, we get
g(x) = (-x^2-x+6)(-ax-3) = ax^3 + (a+3)x^2 + (3-6a)x - 18
So, a+3 = 5 ==> a=2
3-6a = -9
Thus, g(x) = 2x^3 + 5x^2 - 9x - 18 = (-x^2-x+6)(-2x-3)
The remainder when g ( x) is divided by x+2
To find the value of the constants a and b, we will use the fact that f(x) is a factor of g(x). This means that when we divide g(x) by f(x), the remainder should be zero.
Given that f(x) = 6 - x - x^2 and g(x) = ax^3 + 5x^2 + bx + 18, we can divide g(x) by f(x) using long division or synthetic division. Let's use synthetic division here:
Step 1: Arrange the terms of f(x) and g(x) in descending order of the degree of x:
f(x) = -x^2 - x + 6
g(x) = ax^3 + 5x^2 + bx + 18
Step 2: Set up the synthetic division:
```
-1 | 5 -1 b 18
```
Step 3: Perform the synthetic division:
Bring down the first coefficient: 5
Multiply -1 (the divisor) by 5: -5
Add the result to the second coefficient: -1 + (-5) = -6
Multiply -1 by -6: 6
Add the result to the third coefficient: b + 6
Multiply -1 by b + 6: -(b + 6)
Add the result to the fourth coefficient: -(b + 6) + 18 = -b - 6 + 18 = -b + 12
The resulting expression is 5 - 6x - (b + 6)(x + 2). For f(x) to be a factor of g(x), the remainder (-b + 12) should be zero. So, we have:
-b + 12 = 0
-b = -12
b = 12
Therefore, the value of b is 12.
Next, we need to find the value of a. For this, we can substitute the value of b = 12 into the expression we obtained:
g(x) = 5 - 6x - (b + 6)(x + 2)
g(x) = 5 - 6x - (12 + 6)(x + 2)
g(x) = 5 - 6x - 18(x + 2)
Since g(x) = ax^3 + 5x^2 + bx + 18, we can compare the coefficients of x^3 and x^2:
a = -18
Hence, the value of a is -18 and the value of b is 12.