Question

. For x2 + 2x + 5 to be a factor of x' + px + q.
the values of p and q must be
​

Answer 1

Answer:

Let the other factor be x

2

+ax+b. Then

(x

2

+2x+5)(x

2

+ax+b)

≡x

4

+x

3

(2+a)+x

2

(5+b+2a)+x(5a+2b)+5b

≡x

4

+px

2

+q.

Match the coefficients of like powers of x:

For x

3

:2+a=0;∴a=−2

For x:5a+b=0;∴b=5.

For x

2

:5+b+2a=p;∴p=6

For x

0

:5b=q;∴q=25;

or

let y=x

2

so that x

4

+px

2

+q=y

2

+py+q.

Let the roots of y

2

+py+q=0 be r

2

and s

2

.

Since y=x

2

the roots of x

4

+px

2

+q=0 must be ±r,±s.

Now x

2

+2x+5 is a factor of x

4

+px

2

+q; consequently, one pair of roots, say r and s, must satisfy the equation x

2

+2x+5=0. It follows that −r and −s must satisfy the equation x

2

−2x+5=0. Therefore, the other factor must be x

2

−2x+5.

∴(x

2

+2x+5)(x

2

−2x+5)≡x

4

+6x

2

+25≡x

4

+px

2

+q.

∴p=6,q=25;