Find all rational zeros of the polynomial, and write the polynomial in factored form?
P(x)= 2x^3-3x^2-2x+3
2 x ^ 3 - 3 x ^ 2 - 2 x + 3 = 0
( 2 x ^ 3 - 2 x ) - 3 x ^ 2 + 3 = 0
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Remark :
- 3 x ^ 2 + 3 = - ( 3 x ^ 2 - 3 )
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( 2 x ^ 3 - 2 x ) - ( 3 x ^ 2 - 3 ) = 0
2 x ( x ^ 2 - 1 ) - 3 ( x ^ 2 - 1 ) = 0
( x ^ 2 - 1 ) ( 2 x - 3 ) = 0
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Remark :
( x ^ 2 - 1 ) = ( x + 1 ) ( x - 1 )
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( x + 1 ) ( x - 1 ) ( 2 x - 3 ) = 0
2 x ^ 3 - 3 x ^ 2 - 2 x + 3 = ( x + 1 ) ( x - 1 ) ( 2 x - 3 ) = 0
x + 1 = 0
x = - 1
x - 1 = 0
x = 1
2 x - 3 = 0
2 x = 3
x = 3 / 2
The solutions are :
x = - 1
x = 1
and
x = 3 / 2
To find the rational zeros of a polynomial, we can use the Rational Root Theorem. According to this theorem, any rational zero of the polynomial P(x) can be expressed as a fraction p/q, where p is a factor of the constant term (in this case, 3) and q is a factor of the leading coefficient (in this case, 2).
So first, let's find all the factors of the constant term 3. The factors of 3 are 1 and 3. Next, let's find all the factors of the leading coefficient 2. The factors of 2 are 1 and 2.
Using the Rational Root Theorem, we can create a list of all possible rational zeros:
±1/1, ±3/1, ±1/2, ±3/2
Now, to check which of these possible zeros are actual zeros of the polynomial, we can perform synthetic division or use a graphing calculator. By trying each of these values, we can identify the zeros.
By using synthetic division, we find:
For x = 1: The remainder is 0. Therefore, x = 1 is a zero.
Performing synthetic division again, we find:
P(x) = 2x^3-3x^2-2x+3
= (x - 1)(2x^2 - x - 3)
We have factored the polynomial into (x - 1)(2x^2 - x - 3) form.
Now, let's focus on the quadratic factor 2x^2 - x - 3. Factorizing this quadratic equation gives us:
2x^2 - x - 3 = (2x + 3)(x - 1)
So the factored form of the polynomial P(x) = 2x^3-3x^2-2x+3 is:
P(x) = (x - 1)(2x + 3)(x - 1)
Therefore, the rational zeros are x = 1 and x = -3/2.