Factoring the quadratic equation $3x^2-23x+14=0$

Question

I'm having trouble understanding how to factor this equation. Let's go step by step:

1. First I use the sum/product pattern: $$3x^2−2x−21x+14=0$$

2. Then I take the common factors: $$x(3x−2)−7(3x−2)=0$$

3. From this I should obtain: $$(x−7)(3x−2)=0$$

But I do not understand the logic that takes me from point 2) to point 3). What happened?

Answer 1

It is a step justified by distributive property

$$A(B+C) \iff AB+AC$$

in this case we have $A=3x-2$, that is

$$(3x−2)\cdot (x−7) =(3x−2)\cdot x+(3x−2)\cdot (-7)$$

See also the related

• Distributive Law and how it works

Answer 2

Hint: let $u=3x-2$, and take out a factor of $u$.

Answer 3

Alternatively, using the AC method, three times the equation is:

$$9x^2 - 69x + 42 = 0 \implies (3x)^2 - 23(3x) + 42 = 0$$

so you just need to factor $u^2 - 23u + 42 = 0$ where $u =3x$. This gives $(u - 21)(u - 2) = 0 \Rightarrow$ $(3x - 21)(3x - 2) = 0$, or just $(x - 7)(3x - 2) = 0$.