Understanding Engineering Mathematics

In this example division byx−1resultedinaremainderof3/(x− 1 ). In general, when
a polynomialp(x)is divided by a linear factorx−athe remainder can be found directly
by theremainder theorem:

The remainder when a polynomialp(x)is divided byx−ais given by

p(x)

x−a

=q(x)+

r

x−a

wherer=p(a)andq(x)is a polynomial of degree less than that ofp(x).

We can see this by re-writing the above result as

p(x)=(x−a)q(x)+r

and puttingx=ato give

p(a)=r

For example, the remainder whenx^2 +2 is divided byx−1is1^2 + 2 =3, as found
above.

Solution to review question 2.1.9

A.We proceed by repeatedly adding and subtracting terms in the numer-

ator to give us factors containing the denominator, like so:

2 x^3 +x^2 − 6 x+ 9

x+ 2

≡

2 x^3 + 4 x^2 − 4 x^2 +x^2 − 6 x+ 9

x+ 2

≡

2 x^3 (x+ 2 )− 3 x^2 − 6 x+ 9

x+ 2

≡ 2 x^3 −

3 x^2 + 6 x− 9

x+ 2

≡ 2 x^3 −

3 x(x− 2 )− 9

x+ 2

≡ 2 x^3 − 3 x+

9

x+ 2

as above. This method gives you lots of practice in finding factors!

B. From the remainder theorem, the remainder whenp(x)= 4 x^3 − 2 x^2 +

3 x−1 is divided byx−2 is the value ofp(x)whenx=2, i.e.

r= 4. 23 − 2. 22 + 3. 2 − 1 = 29

That is, we can write

4 x^3 − 2 x^2 + 3 x− 1

x− 2

≡ polynomial of degree 2+

29

x− 2