two zeros of cubic polynomial ax^3+3x^2-bx-6 are -1 and -2. find the third zero and values of a and b
if x=-1, then -a + 3 + b - 6 = 0
a - b = -3
if x=-2, then -8a + 12 + 2b - 6 = 0
8a - 2b = 6
4a - b = 3
subtract them:
3a = 6
a = 2
in a-b=-3
2-b = -3
b = 5
so expression is
2x^3 + 3x^2 - 5x - 6
which is (x+1)(x+2)( ?x + ?)
by common sense, the last bracket must be
(2x -3)
so the third root is x=3/2
check:
(x+1)(x+2)(2x-3)
= (x^2 + 3x + 2)(2x-3)
= 2x^3 - 3x^2 + 6x^2 - 9x + 4x - 6
= 2x^3 + 3x^2 - 5x - 6 , all is good
Why did the polynomial go to the circus? Because it wanted to find its missing "zero"!
Now, let's solve this puzzle. Since we know that the polynomial has two zeros at -1 and -2, we can rewrite the polynomial as:
ax^3 + 3x^2 - bx - 6 = a(x + 1)(x + 2)(x - c)
To find the third zero, let's substitute x = -1 and x = -2 into the equation:
a(-1 + 1)(-1 + 2)(-1 - c) = 0
a(-2 + 1)(-2 + 2)(-2 - c) = 0
Simplifying these equations, we get:
a(0)(-c) = 0
a(-1)(-c) = 0
From the first equation, we can deduce that either a = 0 or c = 0. However, if a = 0, then the polynomial would not be cubic. Therefore, we have c = 0.
Substituting c = 0 into the second equation, we get:
a(-1)(0) = 0
This implies that a can be any real number.
So, the third zero of the cubic polynomial is 0, and the values of a and b can be any real numbers.
To find the third zero and the values of a and b, we can use the fact that the sum of the zeros of a cubic polynomial is equal to the opposite of the coefficient of the squared term divided by the coefficient of the cubed term.
Given that two of the zeros are -1 and -2, let's find the third zero:
1. Let's assume the third zero is represented by the variable 'c.'
2. By the zero product property, we know that (x+1)(x+2)(x+c) = 0 represents the given cubic polynomial.
3. Expanding the above expression, we have (x^2+3x+2)(x+c) = 0.
4. Rearranging the terms, we get (x^3 + cx^2 + 3x^2 + 3cx + 2x + 2c) = 0.
5. Combining like terms, we have x^3 + (c+3)x^2 + (3c+2)x + 2c = 0.
Comparing this equation with the given cubic polynomial ax^3 + 3x^2 - bx - 6, we can equate the corresponding coefficients to find the values of a, b, and c.
From the equation, we can deduce the following:
Cubic Term: a = 1 (coefficient of x^3 term)
Squared Term: c+3 = 3 (coefficient of x^2 term)
Linear Term: 3c+2 = -b (coefficient of x term)
Constant Term: 2c = -6 (constant term)
Simplifying these equations, we find:
c = -3 (subtract 3 from both sides of c+3 = 3)
b = 1 (replace c with -3 in 3c+2 = -b)
2c = -6 (substitute c = -3 into 2c = -6)
So, the third zero is -3, and the values of a and b are 1 and -6, respectively.