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This page provides a list of educational videos related to Multiplying Binomials. You can also use this page to find sample questions, apps, worksheets, lessons , infographics and presentations related to Multiplying Binomials.
FOIL method for multiplying binomials example 2
By Khan Academy
FOIL method for multiplying binomials example 2
Example 2: Multiplying a binomial by a polynomial | Algebra I | Khan Academy
By Khan Academy
Example 2: Multiplying a binomial by a polynomial | Algebra I | Khan Academy
Example 1: Multiplying binomials to get a difference of squares | Algebra I | Khan Academy
By Khan Academy
Example 1: Multiplying binomials to get a difference of squares | Algebra I | Khan Academy
Example 3: Modeling with polynomials | Algebra I | Khan Academy
By Khan Academy
Multiplying binomials word problem
Multiply binomials by binomials
By Khan Academy
Sal expresses (x-4)(x+7) as the standard trinomial x^2+3x-28 and discusses how the general product (x+a)(x+b) can be written as x^2+(a+b)x+a*b.
Multiply binomials by binomials
By Khan Academy
Sal expresses the area of a rectangle whose height is x+2 and width is x+3.
Multiply binomials by polynomials
By Khan Academy
Sal finds the values of coefficients a and b that make (2x+4)(5x-9)=ax^2+bx-46 true for all x-values.
Multiply binomials by polynomials
By Khan Academy
Sal expresses the area of a rectangle whose height is y^2-6y and width is 3y^2-2y+1.
Multiply binomials by polynomials
By Khan Academy
Sal expresses the product (10a-3)(5a^2 + 7a - 1) as 50a^3+55a^2-31a+3.
Multiply binomials by polynomials
By Khan Academy
Sal expresses the product (3x+2)(5x-7) as 15x^2-11x-14.
Multiplying binomials and polynomials | Algebra Basics | Khan Academy
By Khan Academy
This instructor in this ten-minute video, Sal Khan, demonstrates how to multiply polynomials. Mr. Khan uses the Paint Program (with different colors) to illustrate his points. Sal Khan is the recipient of the 2009 Microsoft Tech Award in Education. The student or educator may want to open the video to 'full screen' as the instructor is using a black background and the writing is small
Evaluating Logarithms | MathHelp.com
By MathHelp.com
In this example, notice that we have a polynomial divided by a binomial, and our binomial is in the form of an x term minus a constant term, or x – c. In this situation, instead of having to use long division, like we did in the previous lesson, we can divide the polynomials using synthetic division, which is a much more efficient method. Here’s how it works. We start by finding the value of c. Since –c = -3, we know that c = 3. Next, we put the value of c inside a box, so we put the 3 inside a box. It’s very important to understand that the number that goes inside the box always uses the opposite sign as the constant term in the binomial. In other words, since the constant term in the binomial is -3, the number that goes inside the box, is positive 3. Next, we write the coefficients of the dividend, which are 2, -7, 4, and 5. Be very careful with your signs. Now, we’re ready to start our synthetic division. First, we bring down the 2. Next, we multiply the 3 in the box times 2 to get 6, and we put the 6 under the -7. Next, we add -7 + 6 to get -1. Next, we multiply the 3 in the box times -1 to get -3, and we put the -3 under the 4. Next, we add 4 + -3 to get 1. Next, we multiply the 3 in the box times 1 to get 3, and we put the 3 under the 5. Finally, we add 5 + 3 to get 8. Now, notice that we have a 2, -1, 1, and 8 in the bottom row of our synthetic division. These values will give us our answer: the first 3 numbers represent the coefficients of the quotient, and the last number is the remainder. And it’s important to understand that our answer will be one degree less than the dividend. In other words, since our dividend starts with x cubed, and we’re dividing by x, our answer will start with x squared. So our answer is 2x squared – 1x + 1 + 8 over x – 3. Notice that we always use descending order of powers in our quotient. In this case x squared, x, and the constant. Finally, remember that we add the remainder over the divisor, just like we did in the previous lesson on long division, and we have our answer. It’s important to understand that we’ll get the same answer whether we use synthetic division or long division. However, synthetic division is much faster.
McGraw-Hill Math Tutoring | MathHelp.com
By MathHelp.com
This lesson covers the midsegment of triangle. Students learn the following theorems related to parallel lines. If 3 parallel lines form congruent segments along one transversal, then they form congruent segments along every transversal. If a line intersects the midpoint of one side a triangle and is parallel to another side of the triangle, then it intersects the midpoint of the third side of the triangle. If a segment joins the midpoints of two sides of a triangle, then the segment is (a) parallel to the third side of the triangle, and (b) half the length of the third side. Students are then asked to solve problems related to these theorems using Algebra.
