find the roots of polynomial of f(x)=2(x-3)³+16
Answers
Answer:
Step-by-step explanation:
This is an easy step—easy to overlook, unfortunately. If you have a polynomial equation, put all terms on one side and 0 on the other. And whether it’s a factoring problem or an equation to solve, put your polynomial in standard form, from highest to lowest power.
For instance, you cannot solve this equation in this form:
x³ + 6x² + 12x = −8
You must change it to this form:
x³ + 6x² + 12x + 8 = 0
Also make sure you have simplified, by factoring out any common factors. This may include factoring out a −1 so that the highest power has a positive coefficient. Example: to factor
7 − 6x − 15x² − 2x³
begin by putting it in standard form:
−2x³ − 15x² − 6x + 7
and then factor out the −1
−(2x³ + 15x² + 6x − 7) or (−1)(2x³ + 15x² + 6x − 7)
If you’re solving an equation, you can throw away any common constant factor. But if you’re factoring a polynomial, you must keep the common factor.
Example: To solve 8x² + 16x + 8 = 0, you can divide left and right by the common factor 8. The equation x² + 2x + 1 = 0 has the same roots as the original equation.
Example: To factor 8x² + 16x + 8 , you recognize the common factor of 8 and rewrite the polynomial as 8(x² + 2x + 1), which is identical to the original polynomial. (While it’s true that you will focus your further factoring efforts on x² + 2x + 1, it would be an error to write that the original polynomial equals x² + 2x + 1.)
Your “common factor” may be a fraction, because you must factor out any fractions so that the polynomial has integer coefficients.
Example: To solve (1/3)x³ + (3/4)x² − (1/2)x + 5/6 = 0, you recognize the common factor of 1/12 and divide both sides by 1/12. This is exactly the same as recognizing and multiplying by the lowest common denominator of 12. Either way, you get 4x³ + 9x² − 6x + 10 = 0, which has the same roots as the original equation.
Example: To factor (1/3)x³ + (3/4)x² − (1/2)x + 5/6, you recognize the common factor of 1/12 (or the lowest common denominator of 12) and factor out 1/12. You get (1/12)(4x³ + 9x² − 6x + 10), which is identical to the original polynomial.
Step 2. How Many Roots?
A polynomial of degree n will have n roots, some of which may be multiple roots.
How do you know this is true? The Fundamental Theorem of Algebra tells you that the polynomial has at least one root. The Factor Theorem tells you that if r is a root then (x−r) is a factor. But if you divide a polynomial of degree n by a factor (x−r), whose degree is 1, you get a polynomial of degree n−1. Repeatedly applying the Fundamental Theorem and Factor Theorem gives you n roots and n factors.
Descartes’ Rule of Signs
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Descartes’ Rule of Signs can tell you how many positive and how many negative real zeroes the polynomial has. This is a big labor-saving device, especially when you’re deciding which possible rational roots to pursue.
To apply Descartes’ Rule of Signs, you need to understand the term variation in sign. When the polynomial is arranged in standard form, a variation in sign occurs when the sign of a coefficient is different from the sign of the preceding coefficient. (A zero coefficient is ignored.) For example,
p(x) = x5 − 2x3 + 2x2 − 3x + 12
has four variations in sign.
Descartes’ Rule of Signs:
The number of positive roots of p(x)=0 is either equal to the number of variations in sign of p(x), or less than that by an even number.
The number of negative roots of p(x)=0 is either equal to the number of variations in sign of p(−x), or less than that by an even number.
Example: Consider p(x) above. Since it has four variations in sign, there must be either four positive roots, two positive roots, or no positive roots.
Now form p(−x), by replacing x with (−x) in the above:
p(−x) =(−x)5 − 2(−x)3 + 2(−x)2 − 3(−x) + 12
p(−x) = −x5 + 2x3 + 2x2 + 3x + 12
p(−x) has one variation in sign, and therefore the original p(x) has one negative root. Since you know that p(x) must have a negative root, but it may or may not have an