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I'm wrestling with this quadratic and trying to figure out how to factor it:

$$3x^2 - 5x + 2 = 0$$

I know that the product of the last terms of the binomial for an equation equals the third term of the polynomial. Also, the sum of the products of those two numbers should be the middle (second) term of the polynomial.

But what two numbers multiply to give $2$ and added together produce $-5$? How do you use the $3$ in this process?

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6 Answers 6

up vote 5 down vote accepted

Here's a procedure that should help:

To factor $ax^2+bx+c$ first find the product of $a \times c$; in this case, $6$

Then you need to find two numbers that multiply to this value, and add up to $b$; pay attention to the signs of both the product and the sum. In this case, the two numbers are $-2$ and $-3$

Now, the only reason to do this is to split the linear term into two parts that will always let you do the following:$$3x^2 - 5x + 2 = 0$$becomes: $$3x^2 - 3x-2x+ 2 = 0$$and now you can factor by grouping, and then take out a common factor$$3x^2 - 3x -2x+ 2 =(3x^2 - 3x)-(2x-2)=3x(x-1)-2(x-1)=(x-1)(3x-2)$$

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1  
+1 Very nice answer. –  jnh May 6 at 1:09
    
@jnh See my answer for a conceptual presentation of this so-called AC method. –  Bill Dubuque May 6 at 2:01

Hint: $x- 1$ is a factor because $1$ is a solution of the equation..

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Hint $\ $ Reduce to factoring a polynomial that is $\,\rm\color{#c00}{monic}\,$ (lead coeff $=1)$ as follows:

$$\quad\ \ \begin{eqnarray} f &\,=\,& \ \ 3\ x^2-\ 5\ x\ +\ 2\\ \Rightarrow\ 3f &\,=\,& (3x)^2\! -5(3x)+6\\ &\,=\,& \ \ \ \color{#c00}{X^2- 5\ X\,\ +\ 6},\,\ \ X\, =\, 3x\\ &\,=\,& \ \ (X-2)\ (X-\,3)\\ &\,=\,& \ (3x-2)\,(3x-3)\\ \Rightarrow\ f\,=\, 3^{-1}(3f) &\,=\,& \ (3x- 2)\ (\,x\,-1)\\ \end{eqnarray}$$

Remark $\ $ If we denote our factoring algorithm by $\,\cal F,\,$ then the above transformation is simply

$$\cal F f\, = a^{-1}\cal F\, a\,f\quad\,$$

Thus we've transformed by $ $ conjugation $\,\ \cal F = a^{-1} \cal F\, a\ \,$ the problem of factoring non-monic polynomials into the simpler problem of factoring monic polynomials. This is sometimes called the AC method. It works for higher degree polynomials too. As above, we can reduce the problem of factoring a non-monic polynomial to that of factoring a monic polynomial by scaling by a $ $ power of the lead coefficient $\rm\:a\:$ then changing variables: $\rm\ X = a\:x$

$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\rm\: X^2 + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\smash[t]{\overbrace{ac}^{\rm\qquad\ \ \ \ \ {\bf AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\, g(X),\ \ \ X = a\:x \\ \\ \rm\: a^{n-1}(a\:x^n\! + b\:x^{n-1}\!+\cdots+d\:x + c) &\,=\,&\rm\: X^n\! + b\:X^{n-1}\!+\cdots+a^{n-2}d\:X + a^{n-1}c \end{eqnarray}$$ After factoring the monic $\rm\,g(X)\, =\, a^{n-1}f(x),\,$ we are guaranteed that the transformation reverses to yield a factorization of $\rm\:f,\ $ since $\rm\ a^{n-1}$ must divide into the factors of $\rm\ g\ $ by Gauss' Lemma, i.e. primes $\,p\in\rm\mathbb Z\,$ remain prime in $\rm\,\mathbb Z[X],\,$ so $\rm\ p\ |\ g_1(x)\:g_2(x)\,$ $\Rightarrow$ $\,\rm\:p\:|\:g_1(x)\:$ or $\rm\:p\:|\:g_2(x).$

This method also works for multivariate polynomial factorization, e.g. it applies to this question.

For more on the ring-theoretic concepts at the heart of this see this answer.

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Thanks, Bill. Where did you learn how to do all this? –  Alex May 6 at 3:06
2  
@Alex I derived that by reflecting on the algebraic essence of the algorithm used in the high-school AC-method. The ring-theoretic methods at the heart of this deserve to be better-known (see the linked answer). This problem-solving method of "conjugating" problems into simpler problems is ubiquitous, see e.g. Melzak's Bypasses: A Simple Approach to Complexity for many interesting examples. –  Bill Dubuque May 6 at 3:17

Let me offer a different approach. Try to complete the square, by adding and subtracting suitable terms. $3x^2$ begs for an extra $x^2$ to be added to it, so there we go: $$ \begin{aligned} 3x^2−5x+2 &= 4x^2 -4x + 1 - x^2 - x + 1 \\ &= (2x-1)^2 -(x^2- 2x+1) - 3x + 2 \\ &= (2x-1)^2 -(x-1)^2 - (3x - 2) \quad \color{red}{\mbox{<-- $a^2-b^2=(a-b)(a+b)$}}\\ &= (2x-1+x-1)(2x-1-x+1) - (3x-2) \\ &= (3x-2)x-(3x-2) \quad \color{red}{\mbox{<-- factor out $(3x-2)$}}\\ &= (3x-2)(x-1) \end{aligned} $$

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Remark that $$ f(x)=ax^2+bx+c=a(x^2+\frac{b}{a}x+\frac{c}{a})=a(x^2-Sx+P), $$ where S is the sum and P is the product of the zeros of $f$. Hence, $$ 3x^2-5x+2=3(x^2+\frac{-5}{3}x+\frac{2}{3}). $$

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You have an '$a$' term ($3$, the coefficient of $x^2$), so standard factorization won't work. You have to split the middle term as follows:

$3x^2−5x+2=0$

Multiply the coefficient of $x^2$ and the constant, $3(2)=6$. Now, what two number multiply together to get $6$, but add to $-5$? $-3$ and $-2$! (You did this with other quadratics, but for solely $x^2$, the coefficient is $1$ so you can ignore it)

Split the middle term using the numbers from before:

$3x^2-3x-2x+2=0$

Group (Two ways to group, I'll show both with (1) and (2)):

(1) $(3x^2-3x)+(-2x+2)=0$

(2) $(3x^2-2x)+(-3x+2)=0$

Factor normally:

(1) $3x(x-1)-2(x-1)=0$

(2) $x(3x-2)-(3x-2)=0$

Reorder to get a product of binomials:

(1) $(3x-2)(x-1)=0$

(2) $(x-1)(3x-2)=0$

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