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Bad White [126]
3 years ago
7

3x^3-2x^2-147x+98=(ax-c)(bx+d)(bx-d) where a,b,c and d positive integers. Work out the value of a,b,c and d

Mathematics
1 answer:
Alexxx [7]3 years ago
6 0

Expand the right side:

(ax-c)(bx+d)(bx-d)=(ax-c)(b^2x^2-d^2)

=ab^2x^3-b^2cx^2-ad^2x+cd^2

Notice that 98 = 2 * 7 * 7, so we have c=2 and d=7.

Then

-ad^2=-147\implies a=\dfrac{147}{49}=3

and

ab^2=3\implies b^2=1\implies b=1

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