determine the degree and the zeroes of each polynomial function 1.g(x)=(x+2) (x-1) (3x+1)² 2.g(x)=x³(x-1) (x²-1) 3.p(x)=x(x+1)(x²-1) 4.q(x)=(x²-4) (x²-6x+9)
To determine the degree and zeroes of each polynomial function.
step-by-step explanation:
1. g(x) = (x + 2)(x - 1)(3x + 1)²
Degree: The degree of a polynomial is determined by the highest power of x. In this case, the highest power is 2 since (3x + 1)² involves a quadratic term. Therefore, the degree of g(x) is 2.
Zeroes: To find the zeroes, we set g(x) equal to zero and solve for x.
Setting (x + 2) = 0 gives x = -2 as one zero.
Setting (x - 1) = 0 gives x = 1 as another zero.
Setting (3x + 1)² = 0 gives 3x + 1 = 0, which gives x = -1/3 as a repeated zero.
Therefore, the zeroes of g(x) are -2, 1, and -1/3.
2. g(x) = x³(x - 1)(x² - 1)
Degree: The degree of a polynomial is determined by the highest power of x. In this case, the highest power is 3 since x³ is present. Therefore, the degree of g(x) is 3.
Zeroes: To find the zeroes, we set g(x) equal to zero and solve for x.
Setting x³ = 0 gives x = 0 as one zero.
Setting (x - 1) = 0 gives x = 1 as another zero.
Setting (x² - 1) = 0 gives (x - 1)(x + 1) = 0, which gives x = 1 and x = -1 as additional zeroes.
Therefore, the zeroes of g(x) are 0, 1, and -1.
3. p(x) = x(x + 1)(x² - 1)
Degree: The degree of a polynomial is determined by the highest power of x. In this case, the highest power is 3 since x³ is present. Therefore, the degree of p(x) is 3.
Zeroes: To find the zeroes, we set p(x) equal to zero and solve for x.
Setting x = 0 gives x = 0 as one zero.
Setting (x + 1) = 0 gives x = -1 as another zero.
Setting (x² - 1) = 0 gives (x - 1)(x + 1) = 0, which gives x = 1 and x = -1 as additional zeroes.
Therefore, the zeroes of p(x) are 0, -1, and 1.
4. q(x) = (x² - 4)(x² - 6x + 9)
Degree: The degree of a polynomial is determined by the highest power of x. In this case, the highest power is 4 since x² is present in both terms. Therefore, the degree of q(x) is 4.
Zeroes: To find the zeroes, we set q(x) equal to zero and solve for x.
Setting (x² - 4) = 0 gives (x - 2)(x + 2) = 0, which gives x = 2 and x = -2 as zeroes.
Setting (x² - 6x + 9) = 0 gives (x - 3)² = 0, which gives x = 3 as a repeated zero.
Answers & Comments
Answer:
To determine the degree and zeroes of each polynomial function.
step-by-step explanation:
1. g(x) = (x + 2)(x - 1)(3x + 1)²
Degree: The degree of a polynomial is determined by the highest power of x. In this case, the highest power is 2 since (3x + 1)² involves a quadratic term. Therefore, the degree of g(x) is 2.
Zeroes: To find the zeroes, we set g(x) equal to zero and solve for x.
Setting (x + 2) = 0 gives x = -2 as one zero.
Setting (x - 1) = 0 gives x = 1 as another zero.
Setting (3x + 1)² = 0 gives 3x + 1 = 0, which gives x = -1/3 as a repeated zero.
Therefore, the zeroes of g(x) are -2, 1, and -1/3.
2. g(x) = x³(x - 1)(x² - 1)
Degree: The degree of a polynomial is determined by the highest power of x. In this case, the highest power is 3 since x³ is present. Therefore, the degree of g(x) is 3.
Zeroes: To find the zeroes, we set g(x) equal to zero and solve for x.
Setting x³ = 0 gives x = 0 as one zero.
Setting (x - 1) = 0 gives x = 1 as another zero.
Setting (x² - 1) = 0 gives (x - 1)(x + 1) = 0, which gives x = 1 and x = -1 as additional zeroes.
Therefore, the zeroes of g(x) are 0, 1, and -1.
3. p(x) = x(x + 1)(x² - 1)
Degree: The degree of a polynomial is determined by the highest power of x. In this case, the highest power is 3 since x³ is present. Therefore, the degree of p(x) is 3.
Zeroes: To find the zeroes, we set p(x) equal to zero and solve for x.
Setting x = 0 gives x = 0 as one zero.
Setting (x + 1) = 0 gives x = -1 as another zero.
Setting (x² - 1) = 0 gives (x - 1)(x + 1) = 0, which gives x = 1 and x = -1 as additional zeroes.
Therefore, the zeroes of p(x) are 0, -1, and 1.
4. q(x) = (x² - 4)(x² - 6x + 9)
Degree: The degree of a polynomial is determined by the highest power of x. In this case, the highest power is 4 since x² is present in both terms. Therefore, the degree of q(x) is 4.
Zeroes: To find the zeroes, we set q(x) equal to zero and solve for x.
Setting (x² - 4) = 0 gives (x - 2)(x + 2) = 0, which gives x = 2 and x = -2 as zeroes.
Setting (x² - 6x + 9) = 0 gives (x - 3)² = 0, which gives x = 3 as a repeated zero.
therefore, the zeroes of p(x) are 2, -2, and 3.
hope it helps, -eun