Question 110·200 Super-Hard SAT Math Questions·Advanced Math

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x^3 + (b-2)x^2 - (2b+15)x - 15b = 0

In the given equation, $b$ is a positive integer constant. If the equation has exactly two distinct real solutions, which choice could be a solution to the equation?

A cubic with an unknown constant in several coefficients may hide a common factor. Group the terms with and without the constant, then factor to find which roots stay fixed and which root moves. Exactly two distinct real solutions means two factors must give the same root. Watch the sign: $x+b$ gives the root $-b$ , not $b$ .

Hints

Hint 1
Expand the terms containing $b$ first. For example, $(b-2)x^2$ becomes $bx^2-2x^2$ . Can you then put the terms with $b$ in one group and the terms without it in another?
Hint 2
Look for the same quadratic in both groups. Pulling it out as a factor will turn the cubic into smaller pieces whose roots you can compare.
Hint 3
A distinct solution is a different value of $x$ , even if two factors both become zero there. Since $b$ is positive, the root from $x+b$ is negative. Which fixed root could it match?

Step-by-step

Factor and match the repeated root
Step 1
Expand the coefficients containing b
Expand $(b-2)x^2$ and $-(2b+15)x$ so you can separate the $b$ terms from the others:
$0=x^3+bx^2-2x^2-2bx-15x-15b.$
Step 2
Group terms by b
Put the terms without $b$ together and the terms with $b$ together:
$0=(x^3-2x^2-15x)+(bx^2-2bx-15b).$
Step 3
Factor each group
Factor $x$ from the first group and $b$ from the second. They leave the same quadratic:
$0=x(x^2-2x-15)+b(x^2-2x-15).$
Step 4
Pull out the common factor
The repeated quadratic is a common factor, so pull it out:
$0=(x+b)(x^2-2x-15).$
Step 5
Factor the quadratic
Find two numbers that multiply to $-15$ and add to $-2$ : $-5$ and $3$ . Factor the quadratic:
$0=(x+b)(x-5)(x+3).$
Step 6
Find the roots from the factors
A root is a value of $x$ that makes the equation $0$ . A product is $0$ when any factor is $0$ , so set each factor to zero:
$x+b=0,\quad x-5=0,\quad x+3=0.$
Solve the first two equations, and keep the third factor to compare its root with the root of $x+b$ :
$x=-b,\quad x=5,\quad x+3=0.$
Step 7
Make two roots coincide
The root $5$ and the root of $x+3$ are always different. To get exactly two distinct roots, $x+b$ must share one of those roots. Because $b$ is positive, $-b$ cannot be $5$ . To share the root of $x+3$ , the factors $x+b$ and $x+3$ must have the same constant:
$b=3.$
Step 8
Check the root count in Desmos
Type $b=3$ , then graph the original cubic as $y=x^3+(b-2)x^2-(2b+15)x-15b$ . Click its $x$ -intercepts: Desmos shows $(-3,0)$ and $(5,0)$ . The graph touches the axis at $-3$ , but that still counts as only one distinct solution. So $-3$ could be a solution. Choice D.

Question 110·200 Super-Hard SAT Math Questions·Advanced Math

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