Explain how to divide a polynomial by a binomial using an example. (Please refer to the post “How to post equations” in the Main Forum to post your response with correct equations/expressions showing.) Make sure you post the explanation as well as the example.
Well, dividing a polynomial by a binomial can be a little tricky, but don't worry, I'm here to make it fun and easy to understand! Let's jump right into an example:
Suppose we want to divide the polynomial 5x^3 + 7x^2 - 2x + 3 by the binomial x - 2.
To start, we place the dividend (5x^3 + 7x^2 - 2x + 3) under the long division symbol, and the divisor (x - 2) on the left side. We'll be dividing term by term, just like in regular long division.
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x - 2 | 5x^3 + 7x^2 - 2x + 3
Now, we begin by dividing the first term, 5x^3, by x. This gives us 5x^2. We write this result above the line, aligned with the second term of the divisor.
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x - 2 | 5x^3 + 7x^2 - 2x + 3
-5x^3 + 10x^2
_______________
-3x^2
Next, we multiply the divisor (x - 2) by the result we just obtained, which is 5x^2. So, we have 5x^2 * (x - 2), which simplifies to 5x^3 - 10x^2. We write this product underneath the term being subtracted, aligning it with the other subtracted terms.
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x - 2 | 5x^3 + 7x^2 - 2x + 3
-5x^3 + 10x^2
_______________
-3x^2
- (- 3x^2 - 6x)
_______________
4x
Now, we bring down the next term of the dividend, which is -2x, and place it next to the remainder we obtained.
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x - 2 | 5x^3 + 7x^2 - 2x + 3
-5x^3 + 10x^2
_______________
-3x^2
- (- 3x^2 - 6x)
_______________
4x
- 4x + 8
_______________
11
Finally, we bring down the last term of the dividend, which is 3, and place it next to the current remainder.
________________________________________
x - 2 | 5x^3 + 7x^2 - 2x + 3
-5x^3 + 10x^2
_______________
-3x^2
- (- 3x^2 - 6x)
_______________
4x
- 4x + 8
_______________
11
- 11
_______________
Now, since we have no more terms to bring down, and the remainder (-11) is of a lower degree than the divisor, x - 2, we can say that the division is completed. The quotient is 5x^2 - 3x + 4, and the remainder is -11.
So, the polynomial 5x^3 + 7x^2 - 2x + 3 divided by the binomial x - 2 equals 5x^2 - 3x + 4, with a remainder of -11.
Easy peasy, right? Just remember to be patient and take it one term at a time. Happy dividing!
To divide a polynomial by a binomial, you can use the process of long division. Here's a step-by-step explanation with an example:
Example: Divide the polynomial P(x) = 2x^3 + 5x^2 - 4x + 3 by the binomial B(x) = x - 2.
Step 1: Write the polynomial in descending order of powers of the variable x, filling in any missing terms with 0 coefficients. In this example, the polynomial is already in descending order.
P(x) = 2x^3 + 5x^2 - 4x + 3
Step 2: Divide the first term of the polynomial by the first term of the binomial. In this case, divide 2x^3 by x.
2x^3 ÷ x = 2x^2
Step 3: Multiply the result obtained in Step 2 by the entire binomial. In this example, we multiply 2x^2 by (x - 2).
2x^2 * (x - 2) = 2x^3 - 4x^2
Step 4: Subtract the result obtained in Step 3 from the original polynomial. In this example, subtract 2x^3 - 4x^2 from 2x^3 + 5x^2 - 4x + 3.
(2x^3 + 5x^2 - 4x + 3) - (2x^3 - 4x^2) = 5x^2 - 4x + 3 + 4x^2 = 9x^2 - 4x + 3
Step 5: Repeat Steps 2-4 with the new polynomial obtained in Step 4.
9x^2 ÷ x = 9x
9x * (x - 2) = 9x^2 - 18x
(9x^2 - 4x + 3) - (9x^2 - 18x) = 14x + 3
Step 6: Repeat Steps 2-4 with the new polynomial obtained in Step 5.
14x ÷ x = 14
14 * (x - 2) = 14x - 28
(14x + 3) - (14x - 28) = 31
Step 7: At this point, you have obtained the remainder, which is the constant term. The quotient is the sum of the terms obtained in Steps 2, 5, and 6.
Quotient: 2x^2 + 9x + 14
Remainder: 31
Therefore, dividing the polynomial 2x^3 + 5x^2 - 4x + 3 by the binomial x - 2 gives a quotient of 2x^2 + 9x + 14 and a remainder of 31.
To divide a polynomial by a binomial, you can use the method of long division. This method is similar to how you would divide numbers. I will explain the steps using an example:
Example: Divide the polynomial 4x^2 + 7x - 3 by the binomial x + 2.
Step 1: First, make sure both the dividend (the polynomial being divided) and the divisor (the binomial) are in descending order. If they are not, rearrange them accordingly. In our example, both the dividend and the divisor are already in descending order, so we can move on to the next step.
Step 2: Divide the first term of the dividend by the first term of the divisor. In this case, divide 4x^2 by x, which gives us 4x. Write this result on top.
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x + 2 | 4x^2 + 7x - 3
4x
Step 3: Multiply the divisor (x + 2) by the quotient you found in step 2, which is 4x. The product is 4x * (x + 2) = 4x^2 + 8x. Write this product below the dividend.
________
x + 2 | 4x^2 + 7x - 3
4x^2 + 8x
Step 4: Subtract the product from the dividend. This can be done by changing the signs of the terms in the product and then adding.
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x + 2 | 4x^2 + 7x - 3
- (4x^2 + 8x)
____________
15x - 3
Step 5: Bring down the next term from the dividend, which in this case is -3. Write it next to the result obtained so far.
________
x + 2 | 4x^2 + 7x - 3
- (4x^2 + 8x)
____________
15x - 3
- 15
Step 6: Divide the new dividend (15x - 3) by the divisor (x + 2). Similar to step 2, divide 15x by x, which gives us 15. Write this result on top.
________
x + 2 | 4x^2 + 7x - 3
- (4x^2 + 8x)
____________
15x - 3
- 15
_______
15
Step 7: Multiply the divisor (x + 2) by the new quotient you found in step 6, which is 15. The product is 15 * (x + 2) = 15x + 30. Write this product below the current result.
________
x + 2 | 4x^2 + 7x - 3
- (4x^2 + 8x)
____________
15x - 3
- 15
_______
15x + 30
Step 8: Subtract the new product from the current dividend. Change the signs of the terms in the product and then add them to the current dividend.
________
x + 2 | 4x^2 + 7x - 3
- (4x^2 + 8x)
____________
15x - 3
- 15
_______
15x + 30
- (15x + 30)
____________
0
Step 9: Since the result is 0, there is no remainder. The final quotient is the result obtained in step 2 (4x) added to the result obtained in step 6 (15).
Hence, the division of the polynomial 4x^2 + 7x - 3 by the binomial x + 2 is 4x + 15, with no remainder.