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Chapter 6: Problem 129

Factor completely. Assume that variables in exponents represent positive integers. $$ 3(x+1)^{2}+12(x+1)+12 $$

Short Answer

Expert verified
3(x+3)^{2}

Step by step solution

01

- Identify the common factor

First, notice that a common factor of the expression is 3. Factor out the 3 from each term in the expression: 3[(x+1)^{2} + 4(x+1) + 4]
02

- Simplify the quadratic

Next, focus on the quadratic expression inside the brackets. It resembles a standard quadratic form: (x+1)^{2} + 4(x+1) + 4. Let y = (x+1), so the equation becomes: y^{2} + 4y + 4
03

- Factor the quadratic expression

Factor the quadratic equation: y^{2} + 4y + 4 can be written as (y+2)(y+2) or (y+2)^{2}
04

- Substitute back the original variable

Replace y with (x+1) in the factored form: (x+1+2)^{2}, which gives (x+3)^{2}
05

- Write the final factored form

Replace the simplified quadratic back into the original factor: 3[(x+1)^{2} + 4(x+1) + 4] becomes 3(x+3)^{2}

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common Factor
A common factor is a number or term that divides each term in an algebraic expression without leaving a remainder.

In this case, the original expression is 3\((x+1)^{2} + 12(x+1) + 12\).
First, identify that 3 is a common factor across all terms.

By factoring out the 3, the expression simplifies to: \[ 3[(x+1)^{2} + 4(x+1) + 4] \].

This makes the rest of the operations much simpler to handle!
Quadratic Expression
A quadratic expression is an algebraic expression of the form \(ax^2 + bx + c\), where a, b, and c are constants.

In our problem, the expression inside the brackets, \((x+1)^{2} + 4(x+1) + 4\), is a quadratic expression.

It follows the typical quadratic form once we set \(y = (x+1)\). Then, the expression becomes \(y^2 + 4y + 4\).

Recognizing this as a quadratic helps us understand the next steps for factoring.
Factoring Quadratics
Factoring quadratics involves rewriting the quadratic expression as a product of simpler expressions.

For the expression \(y^2 + 4y + 4\), our goal is to express it in the form of \((y + p)(y + q)\).

Notice here that both factors are the same, leading to \((y + 2)(y + 2)\) or \((y + 2)^2\).

This is a neatly factored quadratic form that will help us in the substitution step.
Algebraic Substitution
Algebraic substitution is a method to simplify complex algebraic expressions by temporarily replacing a part of the expression with a single variable.

For our quadratic expression, we set \(y = (x+1)\) to simplify the factoring process.

Then after factoring, we have \((y + 2)^2\).

Once the factoring is complete, we substitute back \(y\) with the original expression \(x+1\), leading us to have \((x+1 + 2)^2\), which simplifies to \((x+3)^2\).
Factoring Completely
To factor completely means to break down an algebraic expression into its simplest building blocks.

After substitution and factoring, we have the simplified term \((x+3)^2\).

The complete factored form of the original expression is found by incorporating the common factor back in. This transforms \(3[(x+1)^{2} + 4(x+1) + 4]\) into \(3(x+3)^2\).

Thus, the entirely factored expression is: \[ 3(x+3)^2 \].

Ensuring each step is checked helps students understand the importance of each stage in the process.

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Most popular questions from this chapter

Factor completely. Assume that variables in exponents represent positive integers. $$ a^{2}+2 a b+b^{2}-c^{2}+6 c-9 $$

Solve. Round any irrational solutions to the nearest thousandth. $$ x^{2}=\frac{1}{100} $$

Solve. Round any irrational solutions to the nearest thousandth. $$ x^{3}+3=3 x^{2}+x $$

Factor completely. Remember to look first for a common factor. If a polynomial is prime, state this. $$ m^{2}-64 $$

Factor completely. Remember to look first for a common factor. If a polynomial is prime, state this. $$ t^{3}+8 t^{2}-t-8 $$
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