multinominal expansion

**fran1942** · February 12th 2011, 02:08 PM

Hello, I have this multinominal and I also have the expanded version of it.
The problem is I cannot understand how the expanded version was achieved.
Can someone please explain it to me. Thanks kindly for any help.

2x^3 + 9x^2 +12x +4

expands to:

(2x+1) (x^2 + 4x +4)

but what is the method ?

**Plato** · February 12th 2011, 02:22 PM

Originally Posted by fran1942

(2x+1) (x^2 + 4x +4) but what is the method ?

The bad news is that it is tedious. Do it term by term.
$(2x)(x^2)+(2x)(4x)+(2x)(4)+(1)(x^2)+(1)(4x)+(1)(4)$

**fran1942** · February 12th 2011, 03:35 PM

thanks kindly for the help.
What I really meant was what method would you follow if you did not know the answer that I provided ?

**Plato** · February 12th 2011, 04:16 PM

Originally Posted by fran1942

thanks kindly for the help.
What I really meant was what method would you follow if you did not know the answer that I provided ?

You follow the method that I gave you.
$(a+b+c+d)(e+f+g)$ is found term by term.
$ae+af+ag+be+bf+bg+ce+cf+cg+de+df+dg$
That is the way it is done.

$(x+1)(x^5+x^3+1)=x^6+x^4+x+x^5+x^3+1$

**fran1942** · February 12th 2011, 04:47 PM

thanks, and how would go in the opposite direction ?

e.g. expanding:

2x^3 + 9x^2 +12x +4

**Plato** · February 12th 2011, 04:50 PM

Originally Posted by fran1942

thanks, and how would go in the opposite direction ?
e.g. expanding: 2x^3 + 9x^2 +12x +4

You would factor the expression not expand it.

**Soroban** · February 12th 2011, 05:02 PM

Hello, fran1942!

Your language is incorrect and confusing . . .

I have this multinominal and I also have the expanded version of it.

The problem is I cannot understand how the expanded version was achieved.

Can someone please explain it to me. Thanks kindly for any help.

$2x^3 + 9x^2 +12x +4$ . This is the expanded version!

. . $\rlap{//////////}\text{expands to: }\;\text{factors to:}$

$(2x+1) (x^2 + 4x +4)$ . This is the factored form!

You are asking how the polynomial was factored, right?

Apply the Rational Roots Theorem and the Remainder Theorem.

**HallsofIvy** · February 13th 2011, 02:22 AM

If this could be factored as " $(ax+ b)(cx^2+ dx+ e)$ " then the leading coefficient would be ac and the constant term would be bd. That is, a must be a factor of the leading coefficient, 2, and b must be a factor of the constant term, 4. Further, if $ax+ b$ is a factor, then $x= -b/a$ (so that ax+ b= 0) would be a zero of the polynomial.

The only possible factors of 2, the only possible values for a, are 2 and 1. The only possible factors of 4, the only possible values of b, are 4, 2, and 1. So the only possible values for $x= b/a$ are 4/2= 2, 2/2= 1, 1/2. Further, since a sum of positive numbers cannot be 0, any possible rational zero of the polynomial must be one of -2, -1, -1/2 and -4.

For
x= -2, the polynomial is $2(-8)+ 9(4)+ 12(-2)+ 4= -16+ 36- 24+ 4= 0$ .
So we get an immediate zero, x= -2, and a factor: x-(-2)= x+ 2.

Now, divide $2x^3 + 9x^2 +12x +4$ by x+ 2:
x+ 2 divides into $2x^3+ 9x^2+ 12x+ 4$ exactly $2x^2+ 5x+ 2$ so we must have
$2x^3 + 9x^2 +12x +4= (x+ 2)(2x^2+ 5x+ 2)$ . Repeating that argument on the last factor, or because it is quadratic, using the quadratic formula to find integer zeros, we find that it cannot be factored further in terms of integer coefficients.

(The quadratic formula gives the roots of $2x^2+ 5x+ 2= 0$ as $\frac{-5\pm\sqrt{25- 8}{4}= \frac{-5}{4}\pm\sqrt{17}}{4}$ so that we can factor into linear terms as $2x^3 + 9x^2 +12x +4= \left(x+ 2\right)\left(x+ \frac{5}{4}+ \frac{\sqrt{17}}{4}\right)\left(x+ \frac{5}{4}- \frac{\sqrt{17}}{4}\right)$ .

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