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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

User Malcooke
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1 Answer

5 votes

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=(147)/(49)=3

and


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

User Oliver Goodman
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