 6.4.1: To divide 16x3  32x2 + 2x + 4 by 4x, divide each term of _________...
 6.4.2: Consider the following long division problem: x + 46x  4 + 2x3. We...
 6.4.3: Consider the following long division problem: 3x  16x3 + 7x2 + 12x...
 6.4.4: In the following long division problem, the first step has been com...
 6.4.5: In the following long division problem, the first two steps have be...
 6.4.6: In the following long division problem, most of the steps have been...
 6.4.7: After performing polynomial long division, the answer may be checke...
 6.4.8: (18x7  9x6 + 20x5  10x4 ) , (2x4 )
 6.4.9: (18a3b2  9a2b  27ab2 ) , (9ab)
 6.4.10: (12a2b2 + 6a2b  15ab2 ) , (3ab)
 6.4.11: (36x4y3  18x3y2  12x2y) , (6x3y3 )
 6.4.12: (40x4y3  20x3y2  50x2y) , (10x3y3 )
 6.4.13: (x2 + 8x + 15) , (x + 5)
 6.4.14: (x2 + 3x  10) , (x  2)
 6.4.15: (x3  2x2  5x + 6) , (x  3)
 6.4.16: (x3 + 5x2 + 7x + 2) , (x + 2)
 6.4.17: (x2  7x + 12) , (x  5)
 6.4.18: (2x2 + x  9) , (x  2)
 6.4.19: (2x2 + 13x + 5) , (2x + 3)
 6.4.20: (8x2 + 6x  25) , (4x + 9)
 6.4.21: (x3 + 3x2 + 5x + 4) , (x + 1)
 6.4.22: (x3 + 6x2  2x + 3) , (x  1)
 6.4.23: (4y3 + 12y2 + 7y  3) , (2y + 3)
 6.4.24: (6y3 + 7y2 + 12y  5) , (3y  1)
 6.4.25: (9x3  3x2  3x + 4) , (3x + 2)
 6.4.26: (2x3 + 13x2 + 9x  6) , (2x + 3)
 6.4.27: (4x3  6x  11) , (2x  4)
 6.4.28: (2x3 + 6x  4) , (x + 4)
 6.4.29: (4y3  5y) , (2y  1)
 6.4.30: (6y3  5y) , (2y  1)
 6.4.31: (4y4  17y2 + 14y  3) , (2y  3)
 6.4.32: (2y4  y3 + 16y2  4) , (2y  1)
 6.4.33: (4x4 + 3x3 + 4x2 + 9x  6) , (x2 + 3)
 6.4.34: (3x5  x3 + 4x2  12x  8) , (x2  2)
 6.4.35: (15x4 + 3x3 + 4x2 + 4) , (3x2  1)
 6.4.36: (18x4 + 9x3 + 3x2 ) , (3x2 + 1)
 6.4.37: f(x) = 8x3  38x2 + 49x  10, g(x) = 4x  1
 6.4.38: f(x) = 2x3  9x2  17x + 39, g(x) = 2x  3
 6.4.39: f(x) = 2x4  7x3 + 7x2  9x + 10, g(x) = 2x  5
 6.4.40: f(x) = 4x4 + 6x3 + 3x  1, g(x) = 2x2 + 1
 6.4.41: x4 + y4 x + y
 6.4.42: x5 + y5 x + y
 6.4.43: 3x4 + 5x3 + 7x2 + 3x  2 x2 + x + 2
 6.4.44: x4  x3  7x2  7x  2 x2  3x  2
 6.4.45: 4x3  3x2 + x + 1 x2 + x + 1
 6.4.46: x4  x2 + 1 x2 + x + 1
 6.4.47: x5  1 x2  x + 2
 6.4.48: 5x5  7x4 + 3x3  20x2 + 28x  12 x3  4
 6.4.49: 4x3  7x2y  16xy2 + 3y3 x  3y
 6.4.50: 12x3  19x2y + 13xy2  10y3 4x  5y
 6.4.51: f(x) = 3x3 + 4x2  x  4, g(x) = 5x3 + 22x2  28x  12, h(x) = 4x + 1
 6.4.52: f(x) = x3 + 9x2  6x + 25, g(x) = 3x3 + 2x2  14x + 5, h(x) = 2x + 4
 6.4.53: ax + 2x + 4 = 3a3 + 10a2 + 6a
 6.4.54: ax  3x + 6 = a3  6a2 + 11a
 6.4.55: Find and interpret f(30). Identify the solution as a point on the g...
 6.4.56: Find and interpret f(70). Identify the solution as a point on the g...
 6.4.57: Rewrite the function by using long division to perform (80x  8000)...
 6.4.58: Rewrite the function by using long division to perform (80x  8000)...
 6.4.59: Explain how to divide a polynomial that is not a monomial by a mono...
 6.4.60: In your own words, explain how to divide by a polynomial containing...
 6.4.61: When performing polynomial long division, explain when to stop divi...
 6.4.62: After performing polynomial long division, explain how to check the...
 6.4.63: When performing polynomial long division with missing terms, explai...
 6.4.64: The idea of supplyside economics is that an increase in the tax ra...
 6.4.65: (6x2 + 16x + 8) , (3x + 2) = 2x + 4
 6.4.66: (4x3 + 7x2 + 8x + 20) , (2x + 4) = 2x2  1 2 x + 3
 6.4.67: (3x4 + 4x3  32x2  5x  20) , (x + 4) = 3x3  8x2 + 5
 6.4.68: Use the TABLE feature of a graphing utility to verify any two divis...
 6.4.69: When performing the division (2x3 + 13x2 + 9x  6) , (2x + 3), I me...
 6.4.70: When performing the division (x5 + 1) , (x + 1), theres no need for...
 6.4.71: Because of exponential properties, the degree of the quotient must ...
 6.4.72: When performing the division (x3 + 1) , (x + 2), the purpose of rew...
 6.4.73: All longdivision problems can be done by the alternative method of...
 6.4.74: Polynomial long division always shows that the answer is a polynomial.
 6.4.75: The long division process should be continued until the degree of t...
 6.4.76: If a polynomial longdivision problem results in a remainder that i...
 6.4.77: (x3n  4x2n  2xn  12) , (xn  5)
 6.4.78: (x3n + 1) , (xn + 1)
 6.4.79: When 2x2  7x + 9 is divided by a polynomial, the quotient is 2x  ...
 6.4.80: Find k so that the remainder is 0: (20x3 + 23x2  10x + k) , (4x + 3).
 6.4.81: Solve: 2x  3 7 4. (Section 4.3, Example 6)
 6.4.82: Write 40,610,000 in scientific notation. (Section 1.7, Example 2)
 6.4.83: Simplify: 2x  4[x  3(2x + 1)]. (Section 1.2, Example 14)
 6.4.84: a. Divide: 5x3 + 6x + 8 x + 2 . b. Find the sum of the numbers in e...
 6.4.85: a. Divide: 3x3  4x2 + 2x  1 x + 1 . b. Find the sum of the number...
 6.4.86: Divide 2x3  3x2  11x + 6 by x  3. Use your answer to factor 2x3 ...
Solutions for Chapter 6.4: Division of Polynomials
Full solutions for Intermediate Algebra for College Students  6th Edition
ISBN: 9780321758934
Solutions for Chapter 6.4: Division of Polynomials
Get Full SolutionsThis expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Intermediate Algebra for College Students, edition: 6. Chapter 6.4: Division of Polynomials includes 86 full stepbystep solutions. Intermediate Algebra for College Students was written by and is associated to the ISBN: 9780321758934. Since 86 problems in chapter 6.4: Division of Polynomials have been answered, more than 53005 students have viewed full stepbystep solutions from this chapter.

Cofactor Cij.
Remove row i and column j; multiply the determinant by (I)i + j •

Companion matrix.
Put CI, ... ,Cn in row n and put n  1 ones just above the main diagonal. Then det(A  AI) = ±(CI + c2A + C3A 2 + .•. + cnA nl  An).

Covariance matrix:E.
When random variables Xi have mean = average value = 0, their covariances "'£ ij are the averages of XiX j. With means Xi, the matrix :E = mean of (x  x) (x  x) T is positive (semi)definite; :E is diagonal if the Xi are independent.

Determinant IAI = det(A).
Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and

Eigenvalue A and eigenvector x.
Ax = AX with x#O so det(A  AI) = o.

Fibonacci numbers
0,1,1,2,3,5, ... satisfy Fn = Fnl + Fn 2 = (A7 A~)I()q A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].

Fundamental Theorem.
The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n  r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.

GramSchmidt orthogonalization A = QR.
Independent columns in A, orthonormal columns in Q. Each column q j of Q is a combination of the first j columns of A (and conversely, so R is upper triangular). Convention: diag(R) > o.

Identity matrix I (or In).
Diagonal entries = 1, offdiagonal entries = 0.

Linearly dependent VI, ... , Vn.
A combination other than all Ci = 0 gives L Ci Vi = O.

Norm
IIA II. The ".e 2 norm" of A is the maximum ratio II Ax II/l1x II = O"max· Then II Ax II < IIAllllxll and IIABII < IIAIIIIBII and IIA + BII < IIAII + IIBII. Frobenius norm IIAII} = L La~. The.e 1 and.e oo norms are largest column and row sums of laij I.

Polar decomposition A = Q H.
Orthogonal Q times positive (semi)definite H.

Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)l has AA+ = 1m.

Schwarz inequality
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.

Simplex method for linear programming.
The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!

Singular matrix A.
A square matrix that has no inverse: det(A) = o.

Skewsymmetric matrix K.
The transpose is K, since Kij = Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.

Solvable system Ax = b.
The right side b is in the column space of A.

Symmetric matrix A.
The transpose is AT = A, and aU = a ji. AI is also symmetric.

Vector space V.
Set of vectors such that all combinations cv + d w remain within V. Eight required rules are given in Section 3.1 for scalars c, d and vectors v, w.