CAT 2023 Slot 1 · Quant · Question 7

Since -2 is a root, the cubic is divisible by \left( x+2 \right), so the other two roots come from the quadratic left behind.

Write the cubic as \left( x+2 \right)\left({x}^{2}+bx+c \right) and expand: {x}^{3}+\left( b+2 \right){x}^{2}+\left( c+2b \right)x+2c.

Matching the constant term, 2c=2, so c=1.

Matching the {x}^{2} coefficient, b+2=2r+1, so b=2r-1.

Checking the x coefficient, c+2b=1+2\left( 2r-1 \right)=4r-1, which matches the given cubic exactly.

So the remaining roots satisfy {x}^{2}+\left( 2r-1 \right)x+1=0.

For these to be real, the discriminant must be non-negative: {\left( 2r-1 \right)}^{2}-4\ge 0.

That means {\left( 2r-1 \right)}^{2}\ge 4, so \left| 2r-1 \right|\ge 2.

Either 2r-1\ge 2, giving r\ge \frac{3}{2}, or 2r-1\le -2, giving r\le -\frac{1}{2}.

Restricting to non-negative integers, the branch r\ge \frac{3}{2} applies, and the smallest such integer is r=2.

Checking at r=2: the quadratic is {x}^{2}+3x+1=0, whose discriminant is 9-4=5>0, so both roots are real.

Shortcut. Notice the constant term is 2 and one root is -2, so the other two roots multiply to -1\times \frac{2}{-2}, that is to 1 — hence the reduced quadratic must be {x}^{2}+bx+1. Only its middle coefficient is unknown, and the real-roots condition \left| b \right|\ge 2 finishes it.