Prove that mx(x^2+2x+3) = x^2-2x-3 has exactly 1 real root if m = 1 & exactly 3 real roots if m =  -2/3
asked Jun 1, 2016 in Other Math Topics by r3dh4r7 (140 points)

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1 Answer

When m=1, x^3+2x^2+3x=x^2-2x-3; x^3+x^2+5x+3=0.

When x=0 the cubic has a value of 3. Because all the signs are plus, when x>0 the cubic stays positive.

When x=-1 the cubic has a value of -2, so there is a root between 0 and -1.

When x<-1 it has negative values because x^3 is always more negative than x^2 and the combined negative values exceed -3 so the whole cubic remains negative.

When m=-2/3, -2x^3/3-4x^2/3-2x=x^2-2x-3; 2x^3/3+7x^2/3-3=0; 2x^3+7x^2-9=0.

This cubic has a root at x=1 because 2+7-9=0.

We can use synthetic division to find the quadratic:

1 | 2 7 0  -9

.....2 2 9   9

.....2 9 9 | 0 = 2x^2+9x+9=0=(2x+3)(x+3). This shows there are three roots: x=1, -3/2, -3.

answered Jun 1, 2016 by Rod Top Rated User (486,860 points)
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