Factoring a quadratic polynomial (absolute beginner level), are both answers correct?

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I'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial:

$$x^2 + 3x - 10$$

And I'm given the task of finding the values of $a$ and $b$ in:

$$(x + a) (x + b)$$

Obviously the answer is: $$(x + 5)(x - 2)$$

However the answer can be also:

$$(x - 2) (x + 5)$$

I just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b$' can be either $5$ or $-2$.

Therefore if a question asks what are the values of '$a$' and '$b$' both the following answers are correct:

Answer $1$
$a = -2$
$b = 5$
or
Answer $2$
$a = 5$
$b = -2$

I'm sure this is a completely obvious question, but I'm just a beginner in this.

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There are 3 best solutions below

3
On BEST ANSWER

.Yes, you are correct. Since $(x+5)(x-2) = (x-2)(x+5) = x^2 + 3x-10$, we note that $a$ and $b$ may either take the values $(5,-2)$ or $(-2,5)$.


I would consider providing just one of the two solutions to be insufficient, since the question itself ask for the values of $a$ and $b$, but nowhere mentions that they are unique. However, any question saying "find the values of $a$ and $b$" is wrong with the word "the" : they are assuming uniqueness of $a$ and $b$, which is not the case.The question as quoted by you includes the word "the" , and this is misleading.

0
On

For commutative property of product we have that

$$(x + 5)(x - 2)=(x - 2)(x + 5)$$

note that also

$$(-x + 2)(-x - 5)$$

is a correct factorization.

0
On

You are right.

$$(x+a)(x+b)=x^2+(a+b)x+ab$$

and by identification with $x^2+3x-10$,

$$\begin{cases}a+b=3,\\ab=-10.\end{cases}$$

This is a non-linear system of equations, and given commutativity of addition and multiplication, it is clear that if $(u,v)$ is a solution, so is $(v,u)$.


Now one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write

$$3a=(a+b)a=a^2+ab=a^2-10$$

which is the original equation (with a sign reversal)

$$a^2-3a-10=0.$$

To be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots.

So there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$.