x2-x-6 is a factor of f(x)=3x3+px2+qx+6. Find the values of the constants p and q and state the zeros of f(x).
f(x) = 3x^3 + px^2 + qx + 6
If x^2-x-6 is a factor, so are (x-3) and (x+2)
that is f(3) = 0 and f(-2) = 0
f(3) = 81 + 9p + 3q + 6 = 0 ----> 3p + q = -29
f(-2) = -24 + 4p - 2q + 6 = 0 -----> 2p - q = 9
add them : 5p = -20
etc, take over ....
Is correct, am very happy
To find the values of the constants p and q, we can use the fact that if x2-x-6 is a factor of f(x), then the remainder when f(x) is divided by x2-x-6 will be zero.
Step 1: Divide f(x) by x2-x-6 using polynomial long division:
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x2 - x - 6 | 3x3 + px2 + qx + 6
- (3x3 - 3x2 - 18x)
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(px2 + qx + 6) + 18x
Step 2: Set the remainder equal to zero and solve for p and q:
(px2 + qx + 6) + 18x = 0
Since the remainder is zero, we have:
px2 + qx + 6 = -18x
Comparing coefficients, we get the following equations:
p = 0
q = -18
So the values of the constants p and q are p = 0 and q = -18.
Step 3: Find the zeros of f(x).
To find the zeros of f(x), we set f(x) equal to zero and solve for x:
3x3 + px2 + qx + 6 = 0
Substituting p = 0 and q = -18, we have:
3x3 - 18x + 6 = 0
The zeros of the function f(x) are the values of x that make f(x) equal to zero. To solve this cubic equation, we can use various methods such as factoring, synthetic division, or plotting the graph.
However, for this specific cubic equation, it is not easy to find the exact zeros analytically. We can approximate the zeros using numerical methods or graphing the equation.
To find the values of the constants p and q, we need to use the fact that x2 - x - 6 is a factor of f(x).
Given that x2 - x - 6 is a factor of f(x), it means that when we divide f(x) by x2 - x - 6, the remainder should be zero.
We can perform polynomial long division to find the quotient and remainder. Here's how we do it:
Step 1: Write f(x) and the divisor x2 - x - 6 as follows:
3x3 + px2 + qx + 6 (dividend)
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x2 - x - 6 | (divisor)
Step 2: Divide the first term of the dividend by the first term of the divisor. In this case, 3x3 ÷ x2 gives us 3x.
3x
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x2 - x - 6
Step 3: Multiply the divisor by the quotient we just found and subtract it from the dividend. This step will cancel out the first term of the dividend and bring down the next term.
3x
____________________
x2 - x - 6 | 3x3 + px2 + qx + 6
- (3x3 - 3x2 - 18x)
____________________
px2 + qx + 18x + 6
Step 4: Repeat steps 2 and 3 with the new expression.
3x + (18x / x2 - x - 6)
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x2 - x - 6 | px2 + qx + 18x + 6
- (px2 - px - 6x)
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(qx + 24x + 6)
Step 5: Divide (qx + 24x + 6) by (x2 - x - 6). This should give us zero remainder.
We need the remainder to be zero, so we can set qx + 24x + 6 equal to zero:
qx + 24x + 6 = 0
To make this equation true for all values of x, the coefficients of all terms should be equal to zero:
q = 0
24 = 0
6 = 0
Since 24 and 6 cannot be equal to zero, the equation qx + 24x + 6 = 0 has no solution for q and p.
Therefore, there are no values of the constants p and q that satisfy the condition x2 - x - 6 being a factor of f(x) = 3x3 + px2 + qx + 6.
As for the zeros of f(x), we can set f(x) equal to zero and solve for x:
3x3 + px2 + qx + 6 = 0
However, since we were unable to find the values of p and q, we cannot determine the zeros of f(x) in this case.