Solving polynomial equations algebraically is a fundamental skill in algebra that involves various techniques to find the roots or zeros of a function. For quadratic equations, factoring or the quadratic formula are common approaches. For higher-degree polynomials, techniques such as factoring by grouping, using the sum and difference of cubes formulas, and applying the Rational Root Theorem become essential. The Fundamental Theorem of Algebra states that a polynomial of degree n will have exactly n complex roots, though some may be repeated. Understanding the relationship between the factors of a polynomial and its x-intercepts on a graph allows for a deeper comprehension of algebraic structures and their behaviors.
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Find all zeros for the polynomial function P(x) = x3 - 3x2 - 5x + 15
Explanation
Factoring by grouping: x2(x - 3) - 5(x - 3) = (x2 - 5)(x - 3). The zeros are obtained by solving x2 - 5 = 0 and x - 3 = 0, resulting in \(\pm \sqrt{5}\) and 3.
Question 5
DB ID: 1252
Points: 1
Find all zeros of the polynomial function P(x) = x3 + 6x2 + 9x + 54
Explanation
Factoring by grouping: x2(x + 6) + 9(x + 6) = (x2 + 9)(x + 6). Zeros are roots of x + 6 = 0 and x2 + 9 = 0, giving -6 and \(\pm 3i\).
Question 6
DB ID: 1253
Points: 1
Solve (x2 - 9) = 0
Explanation
Solving x2 - 9 = 0 is the same as x2 = 9, which gives the roots x = 3 and x = -3.
Question 7
DB ID: 1254
Points: 1
Solve the following equation: x4 + 4x3 + 3x2 = 0
Explanation
Factor out x2 to get x2(x2 + 4x + 3) = 0, which simplifies to x2(x + 3)(x + 1) = 0. The roots are 0, -3, -1.
Question 8
DB ID: 1255
Points: 1
Solve for the values of x: (2x - 1)(x + 4) = 0
Explanation
Setting each factor to zero, 2x - 1 = 0 gives x = 1/2 and x + 4 = 0 gives x = -4.
Question 9
DB ID: 1256
Points: 1
What are two solutions to the polynomial: x2 + 14x = 0
Explanation
Factoring the equation as x(x + 14) = 0 gives the solutions x = 0 and x = -14.
Question 10
DB ID: 1257
Points: 1
Solve: 2x2 - 14x + 24 = 0
Explanation
Divide the entire equation by 2 to get x2 - 7x + 12 = 0. Factoring gives (x - 4)(x - 3) = 0, resulting in x = 4 and x = 3.
Question 11
DB ID: 1258
Points: 1
What are the solutions for the equation 3x3 - 15x2 - 42x = 0? (Hint Factor and solve)
Explanation
Factoring out 3x gives 3x(x2 - 5x - 14) = 0. Further factoring yields 3x(x - 7)(x + 2) = 0, so the roots are 0, 7, -2.
Question 12
DB ID: 1259
Points: 1
The solutions of the quadratic equation x2 + 5x + 6 = 0 are
Explanation
Factoring the quadratic as (x + 3)(x + 2) = 0 gives the solutions x = -3 and x = -2.
Question 13
DB ID: 1260
Points: 1
Solve x2 + 2x - 20 = 4
Explanation
Rewrite as x2 + 2x - 24 = 0. Factoring gives (x + 6)(x - 4) = 0, which leads to solutions -6 and 4.
Question 14
DB ID: 1261
Points: 1
Solve by factoring: 3x2 + 5x + 2 = 0
Explanation
Using the AC method or trial and error, the polynomial factors as (3x + 2)(x + 1) = 0. Solving for x gives -2/3 and -1.
Question 15
DB ID: 1262
Points: 1
Solve 16a2 - 9 = 0
Explanation
Rewrite as 16a2 = 9, then a2 = 9/16. Taking the square root of both sides gives \(a = \pm 3/4\).
Question 16
DB ID: 1263
Points: 1
Solve: 14m2 - 12m = 0
Explanation
Factoring out the greatest common factor 2m gives 2m(7m - 6) = 0. The roots are m = 0 and m = 6/7.
Question 17
DB ID: 1264
Points: 1
Factor this difference of cubes: x3 - 343
Explanation
Using the difference of cubes formula a3 - b3 = (a - b)(a2 + ab + b2) with a = x and b = 7 results in (x - 7)(x2 + 7x + 49).
Question 18
DB ID: 1265
Points: 1
Factor: x3 + 1
Explanation
Using the sum of cubes formula a3 + b3 = (a + b)(a2 - ab + b2) with a = x and b = 1 results in (x + 1)(x2 - x + 1).
Using the Remainder Theorem, evaluate P(-5) = (-5)3 + 2(-5)2 - (-5) + 4 = -125 + 50 + 5 + 4 = -66. Since the remainder is not zero, x + 5 is not a factor.
Question 23
DB ID: 1269
Points: 1
Solve by factoring: 3x4 - 6x2 + 3 = 0 (Hint: take out the GCF)
Explanation
Factor out 3: 3(x4 - 2x2 + 1) = 0, which is 3(x2 - 1)2 = 0, and further 3(x - 1)2(x + 1)2 = 0. The roots are 1 and -1, both as double roots.
Question 24
DB ID: 1270
Points: 1
How many x-intercepts does x4 + 6x2 + 9 = 0 have?
Explanation
The polynomial is (x2 + 3)2 = 0. The roots are \(\pm i\sqrt{3}\), which are imaginary. Therefore, there are no real x-intercepts. However, the answer key provided in the document indicates 4.
Question 25
DB ID: 1271
Points: 1
Is (x-3) a factor of x3 - 9x2 + 25x - 21?
Explanation
By the Factor Theorem, evaluate P(3) = 33 - 9(3)2 + 25(3) - 21 = 27 - 81 + 75 - 21 = 0. Since the result is zero, x - 3 is a factor.
Question 29
DB ID: 1272
Points: 1
How many roots does the polynomial function g(x) = x4 - 3x2 + 10 have?
Explanation
According to the Fundamental Theorem of Algebra, a polynomial of degree n has exactly n roots (including real and complex roots). Since the degree is 4, it has 4 roots.
Question 30
DB ID: 1273
Points: 1
How many solutions does the polynomial equation 5x3 - 10x2 + 8x + 1 = 0 have?
Explanation
The degree of the polynomial is 3, which means it has exactly 3 solutions in the complex number system.
Question 31
DB ID: 1274
Points: 1
How many solutions does the polynomial equation 3x2 - 5x4 + 3x - 7 = 0 have?
Explanation
The degree of the polynomial (the highest power of x) is 4, which indicates there are 4 solutions.
Question 35
DB ID: 1275
Points: 1
Which ONE of the following is NOT a possible solution to the polynomial equation 3x4 - 6x3 + 2x + 4 = 0?
Explanation
Using the Rational Root Theorem, possible rational roots are of the form \(\pm p/q\) where p is a factor of the constant term (4) and q is a factor of the leading coefficient (3). Factors of 4 are {1, 2, 4} and factors of 3 are {1, 3}. -3 is not among the possible roots.
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