Question

Help me on this remainder theorem based question. Useless answers will be reported.

$P_{(x)}=x^3+mx^2+nx+18$ has $(x-a)(x-b)$ as a factor.
Find the value of m and n where a, b, m, n are integers, and where a, b are positive.

Answer 1

Very, veeeery hard question. But it is doable.

By remainder theorem, where x=a or x=b the value will become zero.

(x=a)

$a(a^2+am+n)=-18$

(x=b)

$b(b^2+bm+n)=-18$

Can you see two factors? Those will be factors of -18

By rational zeros theorem, possible values of a, b are factors of 18.

Possible values are 1, 2, 3, 6, 9, 18

Possible choices will be '1, 18', '2, 9', '3, 6'

Where '1, 18' is a pair of a and b

1st equation $1(1+m+n)=-18$

The equation gives $m+n=-19$

2nd equation $18(324+18m+n)=-18$

The equation gives $18m+n=-325$

Solutions are m=-18, n=-1

Where '2, 9' is a pair of a and b

1st equation $2(4+2m+n)=-18$

The equation gives $2m+n=-13$

2nd equation $9(81+9m+n)=-18$

The equation gives $9m+n=-83$

Solutions are m=-10, n=7

Where '3, 6' is a pair of a and b

1st equation $3(9+3m+n)=-18$

The equation gives $3m+n=-15$

2nd equation $6(36+6m+n)=-18$

The equation gives $6m+n=-39$

Solutions are m=-8, n=9

Answer 2

Answer:

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