Problem 51 - Elementary Algebra

Question

Simplify the algebraic expressions for the following problems. $$ 8\left\\{3\left[4 y^{3}+y+2\right]+6\left(y^{3}+2 y^{2}\right)\right\\}-24 y^{3}-10 y^{2}-3 $$

Step-by-Step Solution

Answer

Question: Simplify the following algebraic expression: \(8\{3[4y^3+y+2]+6(y^3+2y^2)\}-24y^3-10y^2-3\). Answer: \(120 y^{3}+86 y^{2}+24 y+45\)

Step 1: Distribute constants to the terms inside the brackets

In this step, we multiply the constants outside the brackets by each term inside them. We start with the inner brackets first. $$ = 8\left\\{3\left[4 y^{3}+y+2\right]+6\left(y^{3}+2 y^{2}\right)\right\\}-24y^{3}-10 y^{2}-3 $$ $$ = 8\left\\{3(4 y^{3}+y+2)+6(y^{3}+2 y^{2})\right\\}-24y^{3}-10 y^{2}-3 $$ $$ = 8\left\\{12 y^{3}+3 y+6+6 y^{3}+12 y^{2}\right\\}-24y^{3}-10 y^{2}-3 $$

Step 2: Simplify the expression inside the curly brackets

Combine like terms within the curly brackets. $$ = 8\left\\{18 y^{3}+12 y^{2}+3 y+6\right\\}-24y^{3}-10 y^{2}-3 $$

Step 3: Distribute the 8 to the terms inside the curly brackets

Multiply 8 by each term within the curly brackets. $$ = (8\cdot18 y^{3})+(8\cdot12 y^{2})+(8\cdot3 y)+(8\cdot6)-24y^{3}-10 y^{2}-3 $$ $$ = 144 y^{3}+96 y^{2}+24 y+48-24 y^{3}-10 y^{2}-3 $$

Step 4: Combine like terms and simplify the expression

Add or subtract the like terms and simplify the resulting expression. $$ = (144 y^{3}-24 y^{3})+(96 y^{2}-10 y^{2})+(24 y)+(48-3) $$ $$ = 120 y^{3}+86 y^{2}+24 y+45 $$ So, the simplified algebraic expression is \(120 y^{3}+86 y^{2}+24 y+45\).

Key Concepts

Distributive Property

The distributive property is a fundamental concept in algebra that allows us to simplify expressions by distributing a multiplier over terms within parentheses. Take for example the expression \(8(3[4y^3+y+2]+6(y^3+2y^2))\). We apply the distributive property by multiplying 8 by each of the terms inside the curly brackets after having already distributed the 3 and 6 over their respective parentheses.

This property makes it easier to work with expressions that have groupings, ensuring that multiplication across addition or subtraction is handled correctly. It involves two key steps: first, distribute within the innermost parentheses, and then, distribute over the outermost brackets or parentheses if there is more than one layer.

Combining Like Terms

When simplifying algebraic expressions, combining like terms is akin to organizing a scattered pile of similar objects. This involves identifying and adding or subtracting terms that have the same variable raised to the same power. In our expression, terms like \(18y^3\) and \(6y^3\) are 'like terms' because they both contain \(y^3\). Once identified, they get combined to simplify the expression to \(144y^3\) after multiplication.

It’s essential for simplification, as it reduces the complexity of algebraic expressions, making them more manageable and easier to understand. By focusing on the common variables and exponents, students can systematically consolidate their expressions.

Elementary Algebra

Elementary algebra is the backbone of understanding higher-level math concepts. It encompasses operations that involve variables, constants, and the use of arithmetic on algebraic expressions. Throughout the process of solving our given problem, we employed basic algebraic operations to manipulate the expression into a simpler form.

In the context of our problem, elementary algebra included distributing, combining like terms, and executing arithmetic operations like addition and subtraction. Mastery of elementary algebra provides a solid foundation for tackling more advanced areas of mathematics and solving a variety of real-world problems.

Constant Multiplication

Constant multiplication is multiplying any term by a constant or fixed number. In our example, the constant 8 is multiplied across each term inside the bracket, following the distributive property, to eliminate the brackets and simplify the expression. This process transformed \(8\cdot18y^3\), \(8\cdot12y^2\), \(8\cdot3y\), and \(8\cdot6\) into \(144y^3\), \(96y^2\), \(24y\), and 48, respectively.

Understanding how to multiply constants with variables and their exponents is crucial because it sets the stage for more complex operations, such as factoring polynomials and solving equations.