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Divide polynomials by polynomials with remainders
By Khan Academy
Sal divides (x^3+5x-4) by (x^2-x+1) using long division.
Divide polynomials by linear binomials with remainders
By Khan Academy
Sal divides (x^3+5x-4) by (x^2-x+1) using long division.
Math Ratios | MathHelp.com
By MathHelp.com
This lesson covers adding decimals. Students learn to add decimals by first lining up the decimal points, then adding the numbers by column. For example, to add 14.2 + 2.86, first line up the decimal points, then add the digits in the hundredths column, to get 0 + 6, or 6, then add the digits in the tenths column, to get 2 + 8, or 10, so write a 0 in the tenths column and carry the 1 to the units column, then add the digits in units column, to get 1 + 4 + 2, or 7, then add the digits in the tens column, to get 1. So 14.2 + 2.86 = 17.06.
Work Word Problems | MathHelp.com
By MathHelp.com
To solve a polynomial inequality, like the one shown here, our first step is to write the corresponding equation. In other words, we simply change the inequality sign to an equals sign, and we have x^2 – 3 = 9 – x. Next, we solve the equation. Since we have a squared term, we first set the equation equal to 0. So we move the 9 – x to the left side by subtracting 9 and adding x to both sides of the equation. This gives us x^2 + x – 12 = 0. Next, we factor the left side as the product of two binomials. Since the factors of negative 12 that add to positive 1 are positive 4 and negative 3, we have x + 4 times x – 3 = 0. So either x + 4 = 0 or x – 3 = 0, and solving each equation from here, we have x = -4, and x = 3. Now, it’s important to understand that the solutions to the equation, -4 and 3, represent what are called the “critical values” of the inequality, and we plot these critical values on a number line. However, notice that our original inequality uses a greater than sign, rather than greater than or equal to sign, so we use open dots on our critical values of -4 and positive 3. Remember that ‘greater than’ or ‘less than’ means open dot, and ‘greater than or equal to’ or ‘less than or equal to’ means closed dot. Now, we can see that our critical values have divided the number line into three separate intervals: less than -4, between -4 and 3, and greater than 3. And here’s the important part. Our next step is to test a value from each of the intervals by plugging the value back into the original inequality to see if it gives us a true statement. So let’s first test a value from the “less than -4” interval, such as -5. If we plug a -5 back in for both x’s in the original inequality, we have -5 squared – 3 greater than 9 minus a -5, which simplifies to 25 – 3 greater than 9 + 5, or 22 greater than 14. Since 22 greater than 14 is a true statement, this means that all values in the interval we’re testing are solutions to inequality, so we shade the interval. Next, we test a value from the “between -4 and 3” interval, such as 0. If we plug a 0 back in for both x’s in the original inequality, we have 0 squared – 3 greater than 9 – 0, which simplifies to 0 – 3 greater than 9, or -3 greater than 9. Since -3 greater than 9 is a false statement, this means that all values in the interval we’re testing are not solutions to inequality, so we don’t shade the interval. Next, we test a value from the “greater than 3” interval, such as 4. If we plug a 4 back in for both x’s in the original inequality, we have 4 squared – 3 greater than 9 – 4, which simplifies to 16 – 3 greater than 5, or 13 greater than 5. Since 13 greater than 5 is a true statement, this means that all values in the interval we’re testing are solutions to inequality, so we shade the interval. Finally, we write the answer that’s shown on our graph in set notation. The set of all x’s such that x is less than -4 or x is greater than 3.
Learn Simplifying Fractions & Equivalent Fractions - [5-4-13]
By Math and Science
Quality Math And Science Videos that feature step-by-step example problems!
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.
