product of two zeros of the polynomial p x = 2 x cube minus
6 X square + 11 x minus 6 is 3 then find its third zero
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Hello, defeated_soldier!
Factor: f(x)=x3−6x2+11x−6
There is the Rational Roots Theorem.
If a polynomial has a rational root, then it is of the form
n
d
. . where n is a factor of the constant term and d is a factor of the leading coefficient.
The constant term is 6 with factors: ±1,±2,±3,±6
The leading coefficient is 1 with factors: ±1
. . Hence, the possible roots are (as Galactus pointed out) are: ±1,±2,±3,±6
Then there is the Factor Theorem.
If f(a)=0, then (x−a) is a factor of f(x).
Get it?
Plug in a number for x ... If it comes out to zero, we've found a factor.
Try x=1:f(1)=13−6⋅12+11⋅1−6=0 . . . Bingo!
. . So, we know that (x−1) is a factor.
Use long (or synthetic) division to get: x3−6x2+11x−6=(x−1)(x2−5x+6)
Then we can factor the quadratic factor:
Factor: f(x)=x3−6x2+11x−6
There is the Rational Roots Theorem.
If a polynomial has a rational root, then it is of the form
n
d
. . where n is a factor of the constant term and d is a factor of the leading coefficient.
The constant term is 6 with factors: ±1,±2,±3,±6
The leading coefficient is 1 with factors: ±1
. . Hence, the possible roots are (as Galactus pointed out) are: ±1,±2,±3,±6
Then there is the Factor Theorem.
If f(a)=0, then (x−a) is a factor of f(x).
Get it?
Plug in a number for x ... If it comes out to zero, we've found a factor.
Try x=1:f(1)=13−6⋅12+11⋅1−6=0 . . . Bingo!
. . So, we know that (x−1) is a factor.
Use long (or synthetic) division to get: x3−6x2+11x−6=(x−1)(x2−5x+6)
Then we can factor the quadratic factor:
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