Finding value of constants given an equation and the equation rewritten with constants

Question

The expression ($x^3$ + $2x^2$)$-$($x^2$ + $6x$) can be rewritten as $(x + a)(x + b)(x + c)$, where $a$, $b$, and $c$ are constants. What is the value of $a + b + c$?

A) -4

B) -1

C) 1

D) 5

How do you solve such a problem given an equation and it rewritten in constant form? Previously seen on SAT.

Answer 1

$$(x^3+2x^2)-(x^2+6x)=x\cdot (x-2)\cdot (x+3)=(x+\color{blue}0)(x\color{blue}{-2})(x+\color{blue}3)$$ Thus $a+b+c=1\implies$ answer C is right.

Alternatively, you can use Vieta's Formulae. These tell us, that you can express the sum of the roots of a monic polynomial as the coefficient of the second term: $$(x^3+2x^2)-(x^2+6x)=x^3+\color{blue}1\cdot x^2-6x+0=0$$

In order to understand why this works try to expand $(x-\alpha)(x-\beta)(x-\gamma)$ and see what happens.

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For further reading, have a look here

Answer 2

Quickest way is to note that $a+b+c$ is the coefficient of $x^2$ in $(x+a)(x+b)(x+c)$. Thus $a+b+c$ must be the coefficient of $x^2$ on the left, hence $2-1=1$.