What Is The Completely Factored Form Of The Expression 16×2 + 8x + 32?

When solving algebraic expressions, you will often be asked to find the factored form of the expression. The factored form of an expression describes how its parts or components (terms) interact with each other.

The terms in an algebraic expression can be variables, constants, products, fractions, powers, and sums (addends). A term can have more than one type of component.

For example, the term x + 2 is a sum of two components: x and 2. The variable x is a component of the term and is also the value that it takes when solving the expression.

Constant values do not change when solving the expression. For example, 8 is a constant value because it does not change when solving 8x = 32. Products and fractions do not change either. Powers and sums may change depending on what number you solve for.

The first factor is (16x + 8)

The second factor is (2x + 4) x 2

The completely factored form of the expression 16×2 + 8x + 32 is (16x + 8)(2x + 4)(32) . Now you can see that the expression is the sum of three quadratic expressions, or equations with two variables.

Quadratic expressions can be very tricky, which is why they have their own name: quadratic equations. A quadratic equation is an equation that has a squared variable or coefficient. For example, x = 5 or x = -5 are both quadratic equations because the variable x appears squared.

There are many ways to solve quadratic equations, and most of them work for any size coefficient or variable so long as you use the correct steps.

Second factor the quadratic expression

Next, factor the second term, 8x + 32. Because this is a numerical term, instead of breaking it up into multiple factors like the first term, you simply count the number of times 8 appears as a factor and then add 1 to that number because there is one extra 8.

There are two of the numbers 8 so the total for this factor is 2+1=3. The number 32 has no other factors so its total is just 1.

Now that you have broken down the entire expression into prime factors, you can re-write the expression in what is called completely factored form. The expression 16×2 + 8x + 32 in completely factored form is 16×2 + 3x + 1.

The second factor is (2x + 4)

The second factor can be called the constant factor. This is because it does not change the value of the product of the exponents and variables.

In this case, the constant factor is 2x + 4. Any number could have replaced the +4, as long as it was not a variable. If it was a variable, then it would have to be a different one.

The difference between this and the first example is that in this case, both variables are multiplied by the same constant factor of 2x. This makes it easier to see that only one x changes in the end product.

The last step is to combine these into one expression: 16×2 + 8x + 32 = (2x + 4)2 * 16×2 + (2x + 4) * 8x + 32 = (8x+8)2 * 16×2

or

16x²+8x+32=8²_×16X^²₊₄

• 8^² × 16X² = 128X²

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  Combine like terms

  When you have finished writing the algebraic expression, you must then combine like terms. Like terms are terms that have the same variable or constant, and have the same number and kind of variables raised to the same powers.

  For example, in the expression 8x + 32, the 8x is a like term with x, since they are both variables with no constant attached. They are also at the same level (8 is an exponent of x) and have the same kind of variable (both are x).

  To combine like terms, you must first add or subtract them if they have the same variable; then, you can multiply or divide them. The goal is to have one variable on top of a constant, such as in 8x + 32 = 10x + 3 = 10x + 30 = 40.

  The completely factored form is (16x + 8)(2x + 4) = 8x(8x + 1) = 8(2+1)+1=8+1=9
  ized>7) Check the final answer by using the product of the factors
  ized>8) Determine if final answer makes sense
  ized>9) Use a calculator to check your work
  Solve Quadratic Equations
  Solve Logarithmic Equations

  Finding the completely factored form of an expression is an important skill to have. Knowing how to do this will help you solve many different types of math problems.

  Quadratic expressions are one of the most common types of equations that you will come across. They are typically found in math related to geometry and solving for unknown values.

  Knowing how to solve quadratic equations is a must-have skill. Luckily, we are here to help!
  Solve Quadratic Equations Solve Logarithmic Equations
  Find Factors
  Find Linear Dependent strong > A pply these skills on these problems: 1) Solve for x: 2×2 + 12x + 8 = 0.
  ized>2) Solve for y: 6y2 − 8y + 24 = 0.
  ized>3) Solve for z: 5z3 − 10z2 + 15z − 30 = 0.
  “
  ized>” An important thing to remember when solving quadratic equations is that if the coefficient of the variable in the squared term is 1, then the variable can be eliminated from the equation.
  ” —OnlineMathHelp Posted October 17th, 2017

  If you were able to follow these steps, then you have solved a quadratic equation! Congratulations! Now that you know how to solve them, let’s look at some more examples. Quadratic equations can be tricky because there are several ways that they can be solved. Luckily, we are here to help! We will give you all of the tips that you need to solve any quadratic equation.Let’s get started by looking at some examples.
  Quadratic Equation Examples
  Here are some examples of quadratic equations that we can solve using our tips above.

  (1) Find all solutions for x such that 2×2 + 12x − 8=0; Solution: By applying Theorem 1 and Factoring The Expression ,we get (2x+4)(x−4)=0; Therefore, there are no real solutions since (-4,-4) satisfies the equation; Thus there are no solutions.(Zero Solution)
  (2) Find all solutions for x such that 3×2−10x+8=0; Solution: By applying Theorem 2 and Factoring The Expression ,we get (3x+1)(x−1)=0; Therefore there is only one real solution since (-1,-1) satisfies the equation; Thus there is one solution.(One Solution )
  (3) Find all solutions for x suchthat 6xy−8y+24=0 ;Solution :By applyingTheorem 3andF actoringTheExpres sion ,weget(6y−4)(y−4)=0 ;Thereforethereareno realso lutionssince( 4,- 4 ) satisfyingtheequation ;Thusthereareno so lutions .( ZeroSolution ) If you were ableTo followthese steps , thenyou have solvedanumberofquadrati calequations !Congratulations !Nowthatyouknowhowtosolvethem ,let’sandlookatsome moreexamplesofquadrati calequations . Let’s first take a look at what a quadratic equation looks like before we try solving it..