Chapter Eleven — Polynomials
The Humongous Book of Algebra Problems
246
11.26 Calculate the product and simplify: (x – y)(x
2
+ 6xy – 9y
2
).
Distribute each term of the left binomial through the right trinomial.
11.27 Calculate the product and simplify: (x + 3y – 1)(2x – y + 7).
Distribute each term of the left trinomial through the right trinomial.
11.28 Calculate the product and simplify: (a – 3b + 2c)(4a – b + 5c).
Distribute each term of the left trinomial through the right trinomial.
Long Division of Polynomials
A lot like long dividing integers
11.29 Calculate (x + 4) ÷ x using long division.
Rewrite the expression as a long division problem, with the divisor outside the
division symbol and the dividend within it.
Even though
both polynomials
have three terms,
it doesn’t change
the way you multiply.
Distribute x, then 3y,
and then –1 through
the right-hand
polynomial and add
everything together:
x(2x – y + 7) + 3y(2x
– y + 7) – 1(2x – y
+ 7).
Before you
try to perform
long division on
polynomials, make
sure you remember
how the process
works with whole
numbers—ip back to
Problems 2.3 and 2.5.
The two techniques
are basically the same,
except with whole
numbers you divide
one DIGIT at a time,
and with polynomials
you divide one
TERM at a time.
The divisor is
what you’re dividing BY and
the dividend is what you’re
dividing INTO.
Chapter Eleven — Polynomials
The Humongous Book of Algebra Problems
247
Divide the leftmost term of the dividend by the divisor: x ÷ x = 1. Write this
number above its like term in the dividend.
Multiply the newly placed number by the divisor: 1 · x = x. Write the opposite of
the result beneath the dividend, once again lining up like terms.
Combine like terms: x – x = 0.
Add the next term of the dividend (+4).
Combine like terms: 0 + 4 = 4.
The degree of the divisor x is greater than the degree of the polynomial below
the horizontal line, so the process of long division is complete.
The number above the division symbol (1) is the quotient and the number
below the horizontal line (4) is the remainder. Write the solution using the
format below.
The number
4 in the dividend
x + 4 has the same
variable as 1 (because
neither of them have
any variables). That
means 1 and 4 are like
terms. When you write
numbers above the
division symbol, line
them up with their
like terms.
Whenever
you multiply a
number on top of the
division symbol by the
number out front, write
the OPPOSITE of that
number (in this case –x
instead of x) beneath
the dividend.
x = x
1
, so x has
degree one. 4 is a
constant and doesn’t
have any variables,
so it has degree zero.
Because 1 > 0, you’re
done dividing.
Chapter Eleven — Polynomials
The Humongous Book of Algebra Problems
248
Note: Problems 11.30–11.31 refer to the quotient (x
2
+ 6x – 9) ÷ (x – 2).
11.30 Calculate the quotient and remainder using long division.
Express the quotient as a long division problem.
Divide the leftmost term of the dividend by the leftmost term of the divisor:
x
2
÷ x = x. Write the answer above the like term 6x in the dividend.
Multiply the newly placed x by each term of the divisor and write the opposites of
the products beneath the dividend.
Combine like terms (6x + 2x = 8x) and add the next term of the dividend (–9).
Divide the leftmost term below the horizontal line by the leftmost term of the
divisor: 8x ÷ x = 8. Write the answer above the division symbol, multiply both
terms of the divisor by 8, and write the opposites of the products below 8x – 9.
Therefore, .
In other words,
x (from the divisor)
times what equals x
2
(from the dividend)?
The answer’s x:
x
.
x = x
2
. If you write
the polynomials from
highest to lowest powers
of x, you’ll always be
dealing with the terms
on the left: “The left
term of the divisor
times what equals
the left term of the
dividend?”
Chapter Eleven — Polynomials
The Humongous Book of Algebra Problems
249
Note: Problems 11.30–11.31 refer to the quotient (x
2
+ 6x – 9) ÷ (x – 2).
11.31 Verify the answer generated by Problem 11.30.
To verify the answer to a long division problem, multiply the divisor by the
quotient and then add the remainder. The result should be equal to the
dividend.
(divisor)(quotient) + remainder = (x – 2)(x + 8) + 7
Multiply the binomials by distributing each term of x – 2 through the quantity
(x + 8), as explained in Problems 11.19–11.24.
Because the result, x
2
+ 6x – 9, is equal to the dividend of Problem 11.30, the
quotient x + 8 and the remainder 7 are correct.
11.32 Calculate (2x
2
– x – 15) ÷ (x
2
+ x – 12) using long division.
Rewrite the expression as a long division problem.
Divide 2x
2
by x
2
to get 2 and write the answer above the like term –15 in the
dividend. Multiply each term of the divisor x
2
+ x – 12 by the newly placed 2
and write the opposite of the results below the corresponding like terms of the
dividend.
The polynomial below the horizontal line has degree 1, which is less than the
degree of the divisor, so the long division process is complete:
The remainder in
Problem 11.30 is just
7, even though you
write the remainder
as part of the
fraction
.
You’re
always
asking, “The
divisor term
with the highest
degree times WHAT
equals the dividend
term with the highest
degree?” In later steps
it changes to, “The
divisor term with the
highest degree times
WHAT equals the
term below the
horizontal line with
the highest
degree?”
.
Chapter Eleven — Polynomials
The Humongous Book of Algebra Problems
250
11.33 Calculate (x
3
+ 4x
2
– 7x + 9) ÷ (x + 1) using long division.
Apply the technique described in Problems 11.29–11.32.
Therefore, .
11.34 Use long division to demonstrate that x + 5 divides evenly into
x
3
+ x
2
– 32x – 60.
Divide x
3
+ x
2
– 32x – 60 by x + 5.
Because the remainder is 0, x + 5 divides evenly into x
3
+ x
2
– 32x – 60.
11.35 Calculate (2x
3
– 4x + 2) ÷ (x – 6) using long division.
Note that the dividend does not contain an x
2
-term. To preserve the polynomial
long division technique, insert the placeholder term 0x
2
into the dividend.
When a
divisor “divides
evenly” into a
dividend, the nal
answer has no
remainder.
That also
means x + 5 is
a FACTOR of
x
3
+ x
2
– 32x – 60. More
on that in Chapter 12
when factoring is
covered in depth.
If the highest
power in the dividend is
x
3
, then you need all the smaller
powers of x in there as well: x
2
, x, and
a constant. If any of those things are
missing, you need to put them in with a
0 coefcient to make sure things
line up properly.
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