I'm doing pre-calculus course at coursera.org and I'm in trouble with this solution $$2x^2 +5x +3 = (2x+3)(x+1)$$
By trial, using ac-method I got stuck: $$ ac = (2)(3) = 6\\ 6 + ? = 5 \Rightarrow~ ? = 5 - 6 = -1 $$
Then, $$2x^2+6x-x+3 = 2x(x+3)-x+3$$
At this point I could not get the answer, any help?
On
In general, $$acx^2+(ad+bc)x+bd=(ax+b)(cx+d).$$ So, in your case, you need to find a set $\{a,b,c,d\}$ such that $$ac=2, ad+bc=5, bd=3.$$
You have $2\times 4=8$ possibilities as $$(a,c)=(1,2),(2,1)$$ $$(b,d)=(1,3),(3,1),(-1,-3),(-3,-1).$$ Now you can find a set $\{a,b,c,d\}$ such that $ad+bc=5.$
So, you'll find that $(a,b,c,d)=(1,1,2,3)$ is what you want. This means you can get $$2x^2+5x+3=(x+1)(2x+3).$$
P.S. : You can use this method for this kind of factorization without any special technique.
On
Complete squares:
$$2x^2+5x+3 = 2(x^2+2\alpha x + \alpha ^2) + (3- 2 \alpha ^2)$$
where $\alpha = \frac{5}{4}$. Then $$(x+\frac{5}{4})^ 2 = \frac{1}{2}\left(\frac{25}{8}-3\right) $$ $$x+\frac{5}{4} = \pm \sqrt{\frac{1}{16}} $$ $$x= \frac{-5}{4} \pm \frac{1}{4} = \frac{-5 \pm 1}{4}$$ $$ x\in \left\{-1, \frac{-3}{2}\right\}$$ Therefore, $$2x^2+5x+3 = (x+1)(2x+3).$$
On
Hint $\ $ Reduce to factoring a polynomial that is $\,\rm\color{#c00}{monic}\,$ (lead coeff $=1)$ as follows:
$$\quad\ \ \begin{eqnarray} f &\,=\,& \ \ 2\ x^2+\ 5\ x\ +\,\ 3\\ \Rightarrow\ 2f &\,=\,& (2x)^2\! + 5(2x)+6\\ &\,=\,& \ \ \ \ \color{#c00}{X^2\!+ 5\ X\ +\,\ 6},\,\ \ X\, =\, 2x\\ &\,=\,& \ \ \,(X+2)\ (X+\,3)\\ &\,=\,& \,\ (2x+2)\,(2x+3)\\ \Rightarrow\ f\,=\, 2^{-1}(2f) &\,=\,& \ \ \ \, (x+1)\,(2x+3)\\ \end{eqnarray}$$
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\ $ (cf. $\rm\color{#0a0}{ac}\,$ below for name = $\rm\color{#0a0}{ac}$-method)
$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\rm\: X^2 + b\:X + \color{#0a0}{ac} =\, g(X),\ \ \ X = a\:x \\ \\ \rm\: a^{n-1}(a\:x^n\! + b\:x^{n-1}\!+\cdots+d\:x + c) &\,=\,&\rm\: X^n\! + b\:X^{n-1}\!+\cdots+a^{n-2}d\:X + a^{n-1}c \end{eqnarray}$$ 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.
Remark $\ $ Those who know university algebra might be interested to know that this works not only for UFDs and GCD domains but also for integrally-closed domains satisfying
$\qquad\qquad$ Primal Divisor Property $\rm\ \ c\ |\ AB\ \ \Rightarrow\ \ c = ab,\ \ a\ |\: A,\ \ b\ |\ B$
Elements $c$ satisfying this are called primal. One easily checks that atoms are primal $\!\iff\!$ prime. Also products of primes are also primal. So "primal" may be viewed as a generalization of the notion "prime" from atoms (irreducibles) to composites.
Integrally closed domains whose elements are all primal are called $ $ Schreier rings by Paul Cohn (or Riesz domains, because they satisfy a divisibility form of the Riesz interpolation property). In Cohn's Bezout rings and their subrings he proved that if $\rm\:D\:$ is Shcreier then so too is $\rm\,D[x],\:$ by using a primal analogue of Nagata's Lemma: an atomic domain $\rm\:D\:$ is a UFD if some localization $\rm\:D_S\:$ is a UFD, for some monoid $\rm\:S\:$ generated by primes. These primal and Riesz interpolation viewpoints come to the fore in a refinement view of unique factorization, which proves especially fruitful in noncommutative rings (e.g. see Cohn's 1973 Monthly survey Unique factorization domains).
In fact Schreier domains can be characterized equivalently by a suitably formulated version of the above "factoring by conjugation" property. This connection between this elementary AC method and Schreier domains appears to have gone unnoticed in the literature.
On
In general $$ax^2+bx+c=a(x-x_1)(x-x_2)$$where $x_1,x_2$ are the roots of equation $$ax^2+bx+c=0$$ in case $a=2,b=5,c=3$ and roots are $$x_{1,2}=\frac{-5\pm\sqrt{5^2-4\cdot2\cdot3}}{2\cdot2}=\frac{-5\pm1}{4},x_1=-1,x_2=-\frac{3}{2}$$ and put these above we get $$2x^2+5x+3=2(x-(-1))(x-(-\frac{3}{2}))=$$ $$=2(x+1)(x+\frac{3}{2})=(x+1)(2x+2\cdot\frac{3}{2})=(x+1)(2x+3)$$
The product of $2,\, 3$ is $6$, as you note. The factors of $6$ which sum to $5$ are $\,2,\, 3.\;$ So we can write $$\begin{align} 2x^2 + 5x + 3 & = \color{blue}{2x^2 +} \underbrace{\color{blue}{2x} + \color{red}{3x}}_{= 5x} \color{red}{+ 3} \\ \\ &= \color{blue}{2x{\bf( x + 1)} } + \color{red}{3{\bf(x+1)}} \\ \\ & = {\bf(x+1)}(2x + 3)\end{align}$$