Division Of Polynomials

17,645 views

Published on

Published in: Education
0 Comments
7 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
17,645
On SlideShare
0
From Embeds
0
Number of Embeds
203
Actions
Shares
0
Downloads
295
Comments
0
Likes
7
Embeds 0
No embeds

No notes for slide

Division Of Polynomials

  1. 1. Division of Polynomials<br />Division of polynomials may look difficult, but it follows the same rules we have already learned for properties of real numbers as well as for division of real numbers.<br />
  2. 2. Divide a Polynomial by a Monomial<br />To divide a polynomial by a monomial, each term is divided by that monomial. Example: (40x2- 15x + 5) ÷ 5 <br /><ul><li>We will first write the problem in fraction format.
  3. 3. We know that when we add or subtract fractions with a common denominator, we merely add or subtract the numerators, and the denominator remains the same.</li></ul>So we can rewrite the problem with each term over that common denominator.<br /><ul><li>Then we simplify each term to get the final result</li></ul>To check: Use distributive property to multiply: <br />5(8x2 – 3x +1) = 40x2 – 15x + 5<br />
  4. 4. Divide a Polynomial by a Monomial<br />Here is another example dividing a polynomial by a monomial where the monomial divisor contains a variable: <br />(12x3 + 24x2 - 8x) ÷ 4x<br /><ul><li>We will first write the problem in fraction format.
  5. 5. Again, we can rewrite the problem with each term over the common denominator.
  6. 6. Then we simplify each term to get the final result</li></ul>To check: Use distributive property to multiply: <br />4x(3x2 + 6x - 2) = 12x3 + 24x2 – 8x<br />
  7. 7. If we are dividing a polynomial by another polynomial, it does not help us to rewrite each term over the divisor. For example, to rewrite (2x2 + x - 3) ÷ (x - 1) as only makes the problem more complicated.<br />As an alternative, we can use long division in the same way it is used for division of real numbers. We write the problem using the division symbol .<br />The first step in dividing is to look at the first term of the divisor (x) and determine how many times it divides into the first term of the dividend, i.e. what would we multiply by x to get 2x2? The result (2x) is written over the x term in the dividend <br />Divide a Polynomial by a Polynomial<br />quotient<br />divisor<br />2x<br />x – 1 2x2 + x – 3<br />dividend<br />
  8. 8. Divide a Polynomial by a Polynomial<br />2x <br />x – 1 2x2 + x – 3<br />2x2 – 2x<br />Next we multiply the partial quotient (2x) by the divisor <br />(x - 1), and place the result below the dividend, lining up the terms <br />Then we subtract that result from the corresponding terms in the dividend and bring down the next term. Note that we are subtracting negative 2x which is the same as adding a positive 2x<br />2x <br />x – 1 2x2 + x – 3<br /> -(2x2 – 2x)<br />3x – 3<br />
  9. 9. Divide a Polynomial by a Polynomial<br />2x + 3<br />x – 1 2x2 + x – 3<br /> -(2x2 – 2x)<br /> 3x - 3<br />Next, we determine how many times x divides into 3x and place the result (3) over the constant term in the dividend.<br />Multiply the 3 by (x – 1) and place the result under the dividend and subtract. We get a remainder of 0 and there are no more terms, so our result is 2x +3.<br />2x + 3<br />x – 1 2x2 + x – 3<br /> -(2x2 – 2x)<br /> 3x - 3<br /> -(3x - 3)<br /> 0<br />
  10. 10. Check the Answer<br />To check, use the FOIL method to multiply.<br /> (x – 1) • ( 2x + 3)<br />First: (x) • (2x) = 2x2<br />Outer: (x) • (3) = 3x<br />Inner: (-1) • (2x) = -2x<br />Last: (-1) • (3) = -3<br />Result: 2x2 + 3x – 2x – 3 = 2x2 + x - 3<br />
  11. 11. Division of Polynomials<br />Another example: (16z3 + 7 – 4z2) ÷ (2z -1) <br />Before we can start the division process we need to rearrange the terms in the dividend, so that they are in descending order of powers. Also note that we have z3 and z2 terms, but no z term. We will place a z term in our dividend with a coefficient of 0 to use as a placeholder. <br />We have rearranged the terms in the dividend, and found our first partial quotient of 8z2, since 2z • 8z2 = 16z3 .<br />Next we multiply 8z2 by (2z -1) and place the result below the dividend lining up the terms with like powers.<br /> 8z2<br />2z – 1 16z3 – 4z2 + 0z +7<br />16z3 – 8z2 <br />
  12. 12. Division of Polynomials<br />Next subtract the first partial product; we get a result of 4z2 and bring down the 0z. <br />Performing the next division: Since 2z • 2z gives us 4z2, we write the 2z in the z term position in the quotient.<br />Now we multiply the 2z by our divisor, write the result below, and perform another subtraction.<br /> 8z2 + 2z<br />2z – 1 16z3 – 4z2 + 0z +7<br /> -(16z3 – 8z2 )<br />4z2 + 0z<br /> 8z2 + 2z<br />2z – 1 16z3 – 4z2 + 0z + 7<br /> -(16z3 – 8z2 )<br /> 4z2 + 0z<br />-(4z2 - 2z)<br /> 2z<br />
  13. 13. Division of Polynomials<br /> 8z2 + 2z + 1<br />2z – 1 16z3 – 4z2 + 0z +7<br /> -(16z3 – 8z2 )<br /> 4z2 + 0z<br />-(4z2 - 2z)<br /> 2z + 7<br />-(2z - 1)<br /> 8<br />Bring down the 7 and perform another division. 2z times 1 will produce the 2z in our new expression. So we continue to build our quotient with a +1, and we multiply that 1 by the divisor and write the result below.<br />We perform another subtraction and get a remainder of 8.<br />
  14. 14. Division of Polynomials<br /> 8z2 + 2z + 1 +<br />2z – 1 16z3 – 4z2 + 0z +7<br /> -(16z3 – 8z2 )<br /> 4z2 + 0z<br />-(4z2 - 2z)<br /> 2z + 7<br />-(2z - 1)<br /> 8<br />Since we have a remainder and no additional terms to bring down, we express that remainder as a fraction by placing the 8 over the divisor of 2z – 1. <br />
  15. 15. Check the Answer<br />To check, we need to multiply (2z -1) (8z2 + 2z + 1 + )<br />We can’t use the FOIL method because our second factor has more than two terms, so we need to multiply each term in the second factor by (2z -1).<br />8z2(2z-1) + 2z(2z-1) + 1(2z-1) + (2z-1)<br />16z3 - 8z2 + 4z2 - 2z + 2z – 1 + 8 Multiply each term<br />16z3 - 4z2 + 7 Combine like terms<br />The result matches our original dividend<br />so our answer is correct.<br />

×