Explanations will vary. The correct answer is 4x2–20xy +25y2.
Now (A + B)2=A2+2AB + B2, so that (A + B)2=A2+B2 just in case 2AB = 0, that is A = 0 or B = 0. The square of the
sum is the sum of the squares in the case (at least) one of the numbers is 0.
Scientific notation is useful with very small numbers.
In order top obtain the entries in the last row, the entries in the upper two rows are being subtracted. They should be
added.
Explanations will vary. The correct answer is -8x3+4x2+ 7x – 5
This is a division of a trinomial by a monomial. Each term in the trinomial must be divided by the monomial, 6.
Cancellation of terms is incorrect.
The correct way to do the division is:
24x2
6–12x
6+6
6 and the correct answer is 4x2–2x + 1.
Answers will vary. One possible answer follows.
Given a polynomial P(x), the remainder of P(x)
x – c is equal to P(c).
Explanations will vary. The correct answer is 8x3–5x2+ 3x + 3
The coefficients of the terms of degree 5 are the same and the coefficients of the terms of degree 4 are not the same.
One solution is -18m11n3.
Expanding (a – b)2 one gets (a – b)2=a2– 2ab + b2 which differs from a2– b2 by the term –2ab and the sign of b2.
The quotient of two monomials may not be a monomial just as the quotient of two integers may not be an integer. For
example, 14x divided by 7xy is equal to 2
y, which is not a monomial because there is a variable in the denominator of
the fraction. (Answers may vary.)
Expanding (x + y)2 one gets (x + y)2=x2+ 2xy + y2 which differs from x2+ y2 by the term 2xy.
Answers will vary. One possible answer follows.
No, I do not agree. The simplest way to evaluate is by direct substitution to get P(0) = –12.
Replace missing terms using 0 as the coefficient of the missing term. That is, 3y5+ 0y4+ 8y3+ 2y2+ 3y+ 5.
The product of a monomial and a binomial is always a binomial. To see this, note that
axn(bxk+cxm) = abxn+k + acxn+m – no canceling is possible.
Answers will vary. One possible answer follows.
Synthetic division can be used. Divide the numerator and denominator by 3 to get
1
3x4 – 5
3x2 + 2
3x – 13
3
x – 2 . The divisor is
now of the form x – c, and synthetic division can be applied.