if 1 is zeroes of the polynomial x³-4x²-7x+10,find the other two
zeroes
Answers
x-1 is a factor of polynomial
On dividing x^3-4x^2-7x+10 by x-1
Factorize the factor imtroduced as quotient on above division.
Then u will get two factors, equate them with 0
And find value of x. The values came are two more zeros.
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Step-by-step explanation:
Polynomials of Degree 3
By Sky Smith
Factoring polynomials helps mathematicians determine the zeros, or solutions, of a function. These zeros indicate critical changes in increasing and decreasing rates and generally simplify the analysis process. For polynomials of degree three or higher, meaning the highest exponent on the variable is a three or greater, factoring can become more tedious. In some instances, grouping methods shorten the arithmetic, but in other cases you may need to know more about the function, or polynomial, before you can proceed further with the analysis.
Analyze the polynomial to consider factoring by grouping. If the polynomial is in the form where the removal of the greatest common factor (GCF) from the first two terms and the last two terms reveals another common factor, you can employ the grouping method. For instance, let F(x) = x³ – x² – 4x + 4. When you remove the GCF from the first and last two terms, you get the following: x²(x – 1) – 4 (x – 1). Now you can pull out (x – 1) from each part to get, (x² – 4) (x – 1). Using the “difference of squares” method, you can go further: (x – 2) (x + 2) (x – 1). Once each factor is in its prime, or nonfactorable form, you are done.
Look for a difference or sum of cubes. If the polynomial has only two terms, each with a perfect cube, you can factor it based on known cubic formulas. For sums, (x³ + y³) = (x + y) (x² – xy + y²). For differences, (x³ – y³) = (x – y) (x² + xy + y²). For example, let G(x) = 8x³ – 125. Then factoring this third degree polynomial relies on a difference of cubes as follows: (2x – 5) (4x² + 10x + 25), where 2x is the cube-root of 8x³ and 5 is the cube-root of 125. Because 4x² + 10x + 25 is prime, you are done factoring.
See if there is a GCF containing a variable which can reduce the degree of the polynomial. For instance, if H(x) = x³ – 4x, factoring out the GCF of “x,” you would get x (x² - 4). Then using the difference of squares technique, you can further breakdown the polynomial into x (x – 2) (x + 2).
Use known solutions to reduce the degree of the polynomial. For example, let P(x) = x³ – 4x² – 7x + 10. Because there is no GCF or difference/sum of cubes, you must use other information to factor the polynomial. Once you find out that P(c) = 0, you know (x – c) is a factor of P(x) based on the "Factor Theorem" of algebra. Therefore, find such a "c." In this case, P(5) = 0, so (x – 5) must be a factor. Using synthetic or long division, you get a quotient of (x² + x – 2), which factors into (x – 1) (x + 2). Therefore, P(x) = (x – 5) (x – 1) (x + 2).
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