Math Problem Statement

Solve the given polynomial equations in part A and use the results to solve the equations in part B.

Solution

Let's work through these polynomial equations step by step. I'll solve Part A first and explain each step clearly.


A. Solve each equation

  1. i. x3=0x - 3 = 0:
    Add 33 to both sides:
    x=3x = 3

  2. ii. x+4=0x + 4 = 0:
    Subtract 44 from both sides:
    x=4x = -4

  3. iii. 3z+2=03z + 2 = 0:
    Subtract 22 and divide by 33:
    z=23z = -\frac{2}{3}

  4. iv. 2z+4=0-2z + 4 = 0:
    Subtract 44, then divide by 2-2:
    z=2z = 2

  5. v. 7x=07x = 0:
    Divide by 77:
    x=0x = 0

  6. vi. 12z=0\frac{1}{2}z = 0:
    Multiply both sides by 22:
    z=0z = 0


B. Use results from Part A

Here we solve each equation in Part B by factoring and using the Zero Product Property. This property states that if ab=0ab = 0, then a=0a = 0 or b=0b = 0.

  1. i. 2(x3)=02(x - 3) = 0:
    Since 202 \neq 0, the solution comes from x3=0x - 3 = 0:
    x=3x = 3

  2. ii. 12(x+4)=0-12(x + 4) = 0:
    Since 120-12 \neq 0, the solution comes from x+4=0x + 4 = 0:
    x=4x = -4

  3. iii. 7x(x+4)=07x(x + 4) = 0:
    Set each factor equal to 00:
    7x=0orx+4=07x = 0 \quad \text{or} \quad x + 4 = 0
    Solve:
    x=0orx=4x = 0 \quad \text{or} \quad x = -4

  4. iv. 12z(3z+2)=0\frac{1}{2}z(3z + 2) = 0:
    Set each factor equal to 00:
    12z=0or3z+2=0\frac{1}{2}z = 0 \quad \text{or} \quad 3z + 2 = 0
    Solve:
    z=0orz=23z = 0 \quad \text{or} \quad z = -\frac{2}{3}

  5. v. (x3)(x+4)=0(x - 3)(x + 4) = 0:
    Set each factor equal to 00:
    x3=0orx+4=0x - 3 = 0 \quad \text{or} \quad x + 4 = 0
    Solve:
    x=3orx=4x = 3 \quad \text{or} \quad x = -4

  6. vi. (3z+2)(2z+4)=0(3z + 2)(-2z + 4) = 0:
    Set each factor equal to 00:
    3z+2=0or2z+4=03z + 2 = 0 \quad \text{or} \quad -2z + 4 = 0
    Solve:
    z=23orz=2z = -\frac{2}{3} \quad \text{or} \quad z = 2


Final Answers:

  • A (part i to vi):
    x=3,x=4,z=23,z=2,x=0,z=0x = 3, \, x = -4, \, z = -\frac{2}{3}, \, z = 2, \, x = 0, \, z = 0

  • B (part i to vi):
    x=3,x=4,x=0or4,z=0or23,x=3or4,z=23or2x = 3, \, x = -4, \, x = 0 \, \text{or} \, -4, \, z = 0 \, \text{or} \, -\frac{2}{3}, \, x = 3 \, \text{or} \, -4, \, z = -\frac{2}{3} \, \text{or} \, 2


Do you want me to clarify any step or show additional explanations?


Related Questions:

  1. How do you solve polynomial equations using factoring?
  2. What is the Zero Product Property and why is it important?
  3. How can you solve higher-degree polynomials using factoring?
  4. What are some tips for solving equations with fractions or coefficients?
  5. How can you verify solutions for polynomial equations?

Tip: Always set each factor of a factored equation equal to 00 to solve polynomial equations efficiently.

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Math Problem Analysis

Mathematical Concepts

Polynomial Equations
Factoring
Zero Product Property
Basic Algebra

Formulas

Zero Product Property: If ab = 0, then a = 0 or b = 0

Theorems

Zero Product Property

Suitable Grade Level

Grades 8-10