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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?

Joe
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Pastille
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    It's due to the fact that $(a+b)c=ac+bc$ – TheBestMagician Sep 03 '21 at 22:04
  • Let me know if you have any questions regarding my answer. – Joe Sep 03 '21 at 22:11
  • Draw a picture: if you have rectangle $a$ units wide and $c$ units long, then the other rectangle that is $b$ units wide and $c$ units long shares a side with the other rectangle. Then just substitute $a = x, b = -7, c = 3x-2$. – Toby Mak Sep 03 '21 at 23:34

3 Answers3

3

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

Joe
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2

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

user
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  • So if I apply the general distributive property to this particular example x⋅(3x−2)−7⋅(3x−2)=0 is B⋅(A)-C⋅(A)=0 which by distributive property can be written as (B-C)⋅A and therefore (x−7)⋅(3x−2)=0 which are the two roots, super clear thanks I just had to apply the general rule to the example! – Pastille Sep 04 '21 at 15:14
  • @Pastille Exactly that’s the underlying rule we are using here! You are welcome! Bye – user Sep 04 '21 at 15:33
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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 $(x - 7)(3x - 2) = 0$.

Toby Mak
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