5-4 Practice Dividing Polynomials Form K

C List the factored form of the polynomial function using your work with Synthetic Division. In this case n1 1 5 4 so the degree of the polynomial is 3 and the polynomial that fi ts these points will be y5ax31bx21cx1d.


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Subtract the bottom binomial from the top binomial.

5-4 practice dividing polynomials form k. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor. X3 x2 14x 27 x 3 3. 1 3 - 4 x2 3 22 2 Determine whether each binomial is a factor of x3 3x2 10x 24.

1x2 - 13x - 482 x 3 2. 2x3 13x2 16x 5 x 5 4. Evaluating a Polynomial Use synthetic division to evaluate fx 5x3 x2 13x 29 when x 4.

Synthetic Division tests zeros. F x x 4 2x3 5x2 8x 4 c fx c rx fx cfx U Example H. Practice 5-4 Form K Divide using long division.

Solution The coefficients of the dividend form the top row of the synthetic division array. 31 scaffolded questions that start relatively easy and end with some real challenges. Zr5 132 16x 5 - x 5 4.

12x2 x - 72 x - 5 3. Here is a set of practice problems to accompany the Dividing Polynomials section of the Polynomial Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University. Now you have four linear equations in four unknowns a b c d.

Name Class Date 8-5 Practice Form K Dividing Polynomials Divide using synthetic division. Practice Dividing Polynomials 5-2 Simplify. 1 20x 4 x3 2x2 4x3 3 20n 4 n3 40n2 10n 5 12x 4 24x3 3x2 6x 7 10n 4 50n3 2n2 10n2 9 x 2 2x 71 x 8 11 n 2 13n 32 n5 13 v 2 2v 89 v 10 15 a 2 4a 38 a 8 17 45p 2 56p 19 9p4 19 10x 2 32x 9 10x 2 21 4r 2 r 1 4r3 23 n 2 4 n 2 25 27b 2 87b 35 3b8.

X 3 x 2 14 x 27 x 3 3. A031b021c0 1d521 or 0a1 0b1 0c1d521. Example 7 Using Synthetic Division Use synthetic division to divide x3 3x2 4x 10 by x 2.

2 x 2 3x - 14 x - 2 9. -6w3z4 - 3w2z5 4w 5z 2w2z 5. F g 4 4 7 f g 4 11.

28d3k2 d k2 - 4dk24dk2-1 7. Given a polynomial and a binomial use long division to divide the polynomial by the binomial. 57 Practice - Divide Polynomials Divide.

Multiply the answer by the divisor and write it below the like terms of the dividend. Free worksheetpdf and answer key on Dividing Polynomials Algebra 2. SOLUTION 4 51 13 29 20 84 388 521 97 359 The remainder is 359.

2 x 3 13 x 2 16 x 5 x 5 4. S3 182 7x - 3 4- 4 - 1 7. X 6.

Bring down the next term of the dividend. 54 Practice - Introduction to Polynomials Simplify each expression. 5-4 Practice Form K Dividing Polynomials Divide using long division.

1x3 5x2 - 3x - 12 x - 1 4. 6 k 3m - 12 k m 2 9 m 3 2k m 3. Practice Form G Dividing Polynomials Divide using long division.

X3 x2 - I4x - 27 - x 3 3. The remainder will be fk. 1 x2-3 12 4 6.

2x2 2x 2. 4a3 2- 8a 2 2a4a-1 6. Jt2 9x 22 x 2 5.

X 2 9 x 22 x 2 5. F 2 7f 10 f 2 8. F g x x 2 5 x 14 x 2 Divide the polynomials.

13x3-x2 - 7x 62 x 2 x5. Substitute for f x and g x. 15 r 10 8- 5 r 40 r 2 2 3 5 r 4 2 2.

2X2 7x - 5 - x l To start divide 9T-2 2x x V2X2 7x - 5 Then multiply 2xx 1 2X2 2x. 6 x 2 4 x 16 2 x. 1 a3 a2 6a 21 when a 4 2 n2 3n 11whenn 6 3 n3 7n2 15n 20 when n2 4 n3 9n2 23n 21whenn5 5 5n4 11n3 9n2 n 5 when n 1 6 x4 5x3 x 13 when x 5 7 x2 9x 23 whenx 3 8 6x3 41x2 32x 11whenx 6 9 x4 6x3 x2 24 when x 6.

Because you are dividing by x 2 write 2 at the top left of the array. X2 9x 22 x. F g x x 7.

Dividing polynomials with long division and synthetic division. 2 x 2 7 x 5 x 1 2. F x 11 X 6 cfx x x 36 4-.

X - 3 9. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor. So to evaluate fx when x k divide fx by x k.

Use these to list out factors use exponents for multiple zeros Example G. So you can conclude from the Remainder Theorem that f4 359. To begin the algorithm bring down the first coefficient.

Set up the division problem. Gx2 4x 16 - 2x - 2 6. Plus model problems explained step by step.

-30x3y 12x 2y2 - 18x2y -6x y 4. F g x x 7 To find f g 4 substitute x 4. Step 2 Substitute the x- and y-values from the four points given in the problem.

X 4 8. Multiply the answer by the divisor and write it. 2x2 7x 5 x 1 To start divide 2 2 2 x x x 2 2 2 1 2 7 5 22 x x x x xx Then multiply 2xx 1 2x2 2x.

In part we found f g x and now are asked to find f g 4.


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