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When factoring a trinomial, in the form $ax^2 + bx + c$, I am told that one can multiply $a$ and $c$ which gives a product whose factors add to $b$.

So if I have $2x^2 + 5x -3$ that gives me $-6$.

Factor $-6$ and add

$1,-6 = -6$

$-1,6 = -6$ (no need to continue, found it)

So $2x^2 + 6x -x -3$

Group and find common factors $(2x^2 + 6x)+(-x -3)$

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

I'm not very good at memorizing formulas but I remember them forever after I understand how to derive them and am able to apply it to more complex things.

My books says $ax^2 + bx + c$ factored can be represented $(px+r)(qx+s)$, and that gives you a product of $pqx^2 + (ps+qr)x + rs$.

So, what we're essentially doing is $rs(pq) =$ set of factors, of which one pair sums to $(ps + qr)$

QUESTION: I don't quite understand where the condition for rs(pq) or ac comes from, so I can't prove it, so I naturally expect to get into trouble with it later. Can someone please help me understand it in pre-algebra terms?

Thanks in advance,

GB

PS, no you aren't doing my home work, I'm doing a self study for taking the CLEP exam, and am planning on being a physics major so I'm wanting to fully understand everything.

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  • $\begingroup$ In pre-algebra terms? You're answering a question in algebra. Is that what you meant to write, or do I misunderstand the American grading system? $\endgroup$ Commented Sep 1, 2015 at 17:28
  • $\begingroup$ I'm sorry I meant pre-calculus $\endgroup$ Commented Sep 1, 2015 at 17:54
  • $\begingroup$ Try taking your example, $2x^2+5x-3$, and double it (to get $4x^2+10x-6$). Now, substitute in $X=2x$, to get $X^2+5X-6$. You know how to solve this — just find things that add up to $5$ and multiply to $-6$. You end up with $X^2+6X-X-6$, right? Now, replace $X$ by $2x$ again and divide by $2$ to get $2x^2+6x-x-3$, which is what your method does. Basically, your method is this but skipping a few steps. $\endgroup$ Commented Sep 1, 2015 at 18:23
  • $\begingroup$ Thanks for the help guys! I really understand what's going on now, I chewed on that for a least an hour before posting. Again, thanks a ton! $\endgroup$ Commented Sep 1, 2015 at 20:05
  • $\begingroup$ Hint: $(x-a)(x-b)=x^2-(a+b)x+ab$. $\endgroup$ Commented Sep 1, 2015 at 20:46

3 Answers 3

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This method is just a clever way to set things up so that we can factor by grouping. If $a\ne0$ and we can write $ac=\alpha\beta$ such that $\alpha+\beta=b,$ then we have: $$\begin{align}ax^2+bx+c &= ax^2+(\alpha+\beta)x+\frac{\alpha\beta}a\\ &= ax^2+\alpha x+\beta x+\frac{\alpha\beta}a\\ &= ax\left(x+\frac\alpha a\right)+\beta\left(x+\frac\alpha a\right)\\ &=\left(ax+\beta\right)\left(x+\frac\alpha a\right)\end{align}$$ Alternately, we could rewrite $b=\beta+\alpha$ and get the (possibly different) factorization $$ax^2+bx+c=\left(ax+\alpha\right)\left(x+\frac\beta a\right)$$ by similar manipulations. Alternately, we could pull out the factor of $a$ and get $$ax^2+bx+c=a\left(x+\frac\alpha a\right)\left(x+\frac\beta a\right)$$ in either case.

Now, in theory, we can always do this. After all, if $a,b,c$ are real numbers and $a$ is non-zero, then we should be able to rewrite $$ax^2+bx+c=(px+q)(rx+s)$$ for some $p,q,r,s.$ Expanding the right hand side shows that $a=pr,$ $b=ps+qr,$ and $c=qs,$ from which we see that $ac=prqs=psqr,$ so $ps,qr$ are factors of $ac$ that sum to $b.$ In practice, however, this may be anything but straightforward.

For example, let's say we're given the polynomial $16x^2-24x+43.$ Now, if (by some miracle) we realize that by putting $\alpha=-12+4i\sqrt{34}$ and $\beta=-12-4i\sqrt{34},$ we will then have (readily) that $$\alpha+\beta=b$$ and (not so readily) that $$\alpha\beta=144+16\cdot34=688=ac,$$ then we can factor as above, but how in the world would we just notice that?

Ultimately, it is usually better to proceed by another method.

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When $a=1$, you can factor $x^2+bx+c=(x+\alpha)(x+\beta)=x^2+(\alpha+\beta)x+\alpha\beta$, and the rule is obvious: factor the independent coefficient $c=\alpha\beta$ and sum to get the linear coefficient $b=\alpha+\beta$.

When $a\ne1$, we can modify the procedure by multiplying by $a$ to yield

$$a^2x^2+abx+ac,$$ and solve for $z=ax$ instead of $x$:

$$a^2x^2+abx+ac=z^2+bz+ac.$$

We are now back in the first case and we apply the rule with $ac$ as the independent coefficient and $b$ the linear coefficient.

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$f(x)=ax^2 + bx + c = a(x^2 + \frac ba x + \frac ca)$ assuming $a\neq0$

Continuing.. $f(x)=a(x^2+\frac ba x+ \frac{b^2}{4a^2} -\frac{b^2}{4a^2}+\frac ca)$

$f(x)=a[(x+\frac{b}{2a})^2-(\frac{\sqrt(b^2-4ac)}{2a})^2]$ ...... (Milestone 1)

Now any polynomial of the form $g(x,y)=x^2-y^2$ can be simplified as below -

$g(x,y)=x^2-xy+xy-y^2=x(x-y)+y(x-y)=(x+y)(x-y)$

Using this result and continuing from milestone 1, we have

$f(x)=a(x+\frac{b+\sqrt(b^2-4ac)}{2a})(x+\frac{b-\sqrt(b^2-4ac)}{2a})$

which is of the form $(px+r)(qx+s)$ if looked at as $pq(x+\frac rp)(x+\frac sq)$

Thus, it can be seen that one can always find the set of numbers ${p,q,r,s}$ for a trinomial $ax^2+bx+c$ to be expressed as $(px+r)(qx+s)$

Once this is done, we get the condition for $rs(pq)$.

Hope this helped.

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