Advanced Algebra – Unit 3B Polynomials Test Review Answer Key Advanced Algebra – Unit 3B Polynomials Test Review Answer Key
Advanced Algebra – Unit 3B Polynomials Test Review Answer KeyAdvanced Algebra – Unit 3B Polynomials Test Review Answer KeyAdvanced Algebra – Unit 3B Polynomials Test Review Answer KeyIn advanced algebra, polynomials are a fundamental concept that students must understand in order to solve complex equations and functions. Unit 3B of the advanced algebra curriculum focuses on polynomials, including topics such as factoring, operations with polynomials, and solving polynomial equations. To help students prepare for the upcoming test on polynomials, we have provided an answer key to the test review below.Question 1:Factor the following polynomial completely: $4x^2 + 12x + 9$Answer:$4x^2 + 12x + 9 = (2x + 3)(2x + 3) = (2x + 3)^2$Question 2:Simplify the expression: $(x^2 + 3x – 4) + (2x^2 – 5x + 7)$Answer:$(x^2 + 3x – 4) + (2x^2 – 5x + 7) = 3x^2 – 2x + 3$Question 3:Solve the equation: $2x^2 – 7x + 3 = 0$Answer:To solve the equation, we can use the quadratic formula: $x = \frac{-b ± \sqrt{b^2 – 4ac}}{2a}$In this case, $a = 2, b = -7, c = 3$Substitute the values into the formula: $x = \frac{7 ± \sqrt{(-7)^2 – 4*2*3}}{2*2}$Simplify: $x = \frac{7 ± \sqrt{49 – 24}}{4}$$= \frac{7 ± \sqrt{25}}{4}$$= \frac{7 ± 5}{4}$$x_1 = \frac{7 + 5}{4} = 3$$x_2 = \frac{7 – 5}{4} = \frac{1}{2}$Question 4:Factor the following polynomial completely: $6x^3 – 27x^2 + 18x$Answer:$6x^3 – 27x^2 + 18x = 3x(2x – 3)(x – 2)$Question 5:Find the zeros of the polynomial: $x^4 – 8x^3 + 15x^2$Answer:To find the zeros, we set the polynomial equal to zero and solve:$x^4 – 8x^3 + 15x^2 = 0$Factor out an x^2: $x^2(x^2 – 8x + 15) = 0$Factor the quadratic: $x^2(x – 3)(x – 5) = 0$This gives us zeros at $x = 0, x = 3, x = 5$Question 6:Simplify the expression: $(3x^2 + 5x – 2)(x^2 – 4x + 1)$Answer:To simplify the expression, we can expand using the distributive property:$(3x^2 + 5x – 2)(x^2 – 4x + 1) = 3x^4 – 12x^3 + 3x^2 + 5x^3 – 20x^2 + 5x – 2x^2 + 8x – 2$Combine like terms: $3x^4 – 7x^3 – 19x^2 + 13x – 2$Question 7:Solve the equation: $x^3 – 6x^2 + 11x – 6 = 0$Answer:To solve the equation, we can use synthetic division or trial and error to find the zeros:$x = 1$ is a zero of the polynomial.Perform synthetic division with $(x – 1)$:$1 | 1 -6 11 -6$$|___ 1 -5 6$This leaves us with the reduced polynomial: $x^2 – 5x + 6$Factor the reduced polynomial: $(x – 2)(x – 3) = 0$The zeros are $x = 1, x = 2, x = 3$Question 8:Factor the following polynomial completely: $4x^4 – 16x^2$Answer:$4x^4 – 16x^2 = 4x^2(x^2 – 4) = 4x^2(x + 2)(x – 2)$Question 9:Find the sum of the zeros of the polynomial: $2x^3 – 7x^2 + 3x – 1$Answer:To find the sum of the zeros, we can use Vieta’s formulas which state that the sum of the zeros is equal to the opposite of the coefficient of $x^2$ divided by the leading coefficient.In this case, the sum of the zeros = $\frac{7}{2} = 3.5$Question 10:Simplify the expression: $\frac{x^3 – 2x^2 – 8x}{x^2 – 4}$Answer:To simplify the expression, we can divide the numerator by the denominator. We can use long division or synthetic division to divide $x^3 – 2x^2 – 8x$ by $x^2 – 4$.$x^3 – 2x^2 – 8x = (x – 4)(x^2 + 2x) = x(x – 4)(x + 2)$Therefore, $\frac{x^3 – 2x^2 – 8x}{x^2 – 4} = x(x – 4)(x + 2)$By going through these practice questions, students can better prepare for the upcoming test on polynomials in advanced algebra. Understanding how to factor polynomials, solve polynomial equations, and simplify expressions will be crucial for success on the test. Reviewing key concepts and practicing similar problems will help reinforce students’ understanding of polynomials and improve their problem-solving skills in advanced algebra. Good luck on the test!

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