Why, my result has the opposite sign to the correct result?

Question

I found this exercise of factoring:

$$5a^2 + 3a - 2$$ is a trinomial of the form:

$$ax^2 + bx + c = \frac{(ax + p)(ax + q)}{a}$$

I made this:

$$3 = p + q$$

$$-10 = pq$$

then, I need to get the values of $p$ and $q$.

$$p = -\frac{10}q$$

replace:

$$3 = -\frac{10}q + q$$

$$q^2 -3q - 10 = 0$$

$$(q - 5)(q + 2)$$ That is two possible values of $q$

I need to check the correct value now:

$$3 = p - 5\implies p = -8$$ but $8 \times 5 \neq -10$, so the correct value of $q$ is $2$.

Replace to check if is true:

$$3 = p - 2\implies p = 1$$ but $1 \times -2 \neq -10$

In the end, the correct values impressively were $5$ and $2$, which I had obtained in the equation of $q$, but with a changed sign. $$p = 5, q = -2$$

So, what am I doing wrong factoring this expression? Is it a property that the result was that? or was it coincidence

PD: the total factoring was: $(a + 1)(5a - 2)$

Answer 1

The mistake is when you stated that $(q-5)(q+2)=0$ has solutions $q=-5,q=2$.

Instead, you have $$q-5=0\quad\text{and}\quad q+2=0$$ giving $q=5$ and $q=-2$ and so substituting gives $$3=p+5\implies p=-2\implies pq=-2(5)=-10$$ and $$3=p-2\implies p=5\implies pq=5(-2)=-10$$ as desired.

Answer 2

the total factoring was: $\;(a + 1)(5a - 2)$

You could get that directly by finding the roots using the quadratic formula $\,\dfrac{-3 \pm \sqrt{49}}{10}=\begin{cases}-1 \\ 2/5\end{cases}\,$.

Then the quadratic factors as $5\big(a-(-1)\big)\left(a - \dfrac{2}{5}\right)=(a+1)(5a-2)\,$.

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[ EDIT ] In the general case $\,ax^2 + bx + c = \dfrac{(ax + p)(ax + q)}{a}\,$ the factorization follows directly from the quadratic formula as well:

$$ \begin{align} ax^2 + bx + c = a(x-x_1)(x-x_2) &= a\left(x-\frac{-b - \sqrt{\Delta}}{2a}\right)\left(x-\frac{-b + \sqrt{\Delta}}{2a}\right) \\[10px] &= \frac{\left(ax + \cfrac{b + \sqrt{\Delta}}{2}\right)\left(ax + \cfrac{b - \sqrt{\Delta}}{2}\right)}{a} \end{align} $$

Answer 3

Note that $$5a^2 + 3a - 2= (5a+p)(a+q)=5a^2+(p+5q)a+pq$$

Thus $p+5q=3$ and $ pq=-2$

which implies $q=1$ and $p=-2$

Therefore the factorization is
$$5a^2 + 3a - 2=(5a+p)(a+q)=(5a-2)(a+1)$$

Answer 4

I'm satisfied with TheSimplyFire's answer for your question. Also look at the equation you set up: $$ax^2 + bx + c = \frac{(ax + p)(ax + q)}{a}$$ Now look at the coefficients of $x$: It is $a$ in the both factors of $x$ on RHS. Thus, $p$ and $q$ should be interchangeable and factors belong to only one of two factor pairs of $-10$. That means, $p\times(q)=2\times(-5)$ or $p\times(q)=5\times(-2)$ or $q\times(p)=2\times(-5)$ or $q\times(p)=5\times(-2)$.

Note: Your equation setup can be consider as general form. However, if $a$ in $ax^2 + bx + c$ is a prime number like $5$, you do not have to divide the RHS of the equation again by $a$. Instead, you can setup it as: $$ax^2 + bx + c = (ax + p)(x + q)$$ This is because any prime number $p$ has one distinctive factor pair: $p\times1$.