Question

Factorize `x^4–3x^3+4x^2-8` ?

Answer 1

`x^4-3x^3+4x^2-8 = (x+1)(x-2)(x^2-2x+4)`

`color(white)(x^4-3x^3+4x^2-8) = (x+1)(x-2)(x-1-sqrt(3)i)(x-1+sqrt(3)i)`

Explanation:

Given:

`f(x) = x^4-3x^3+4x^2-8`

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-8` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-4, +-8`

We find:

`f(-1) = 1+3+4-8 = 0`

So `x=-1` is a zero and `(x+1)` a factor:

`x^4-3x^3+4x^2-8 = (x+1)(x^3-4x^2+8x-8)`

We also find that `x=2` is a zero of the remaining cubic, so `(x-2)` is a factor:

`x^3-4x^2+8x-8 = (x-2)(x^2-2x+4)`

The remaining quadratic has a negative discriminant, so no real zeros or linear factors with real coefficients.

We can factor it with complex coefficients by completing the square:

`x^2-2x+4 = x^2-2x+1+3`

`color(white)(x^2-2x+4) = (x-1)^2-(sqrt(3)i)^2`

`color(white)(x^2-2x+4) = ((x-1)-sqrt(3)i)((x-1)+sqrt(3)i)`

`color(white)(x^2-2x+4) = (x-1-sqrt(3)i)(x-1+sqrt(3)i)`

Answer 2

`x^4-3x^3+4x^2-8 = (x^2-2x+4)(x-2)(x+1)`

Explanation:

For illustration, here's what happened when I thought this was a complicated quartic and started a full blown quartic solution (not recommended unless you really need to do it)...

Given:

`x^4–3x^3+4x^2-8`

Let us get this into depressed quartic form with a Tschirnhaus transformation...

`256(x^4–3x^3+4x^2-8)`

`=256x^4-768x^3+1024x^2-2048`

`=(4x-3)^4+10(4x-3)^2+168(4x-3)-1715`

`=t^4+10t^2+168t-1715`

where `t = 4x-3`

Next note that since the quartic in `t` has no `t^3` term, it must factor as a pair of quadratics with opposite middle coefficient:

`t^4+10t^2+168t-1715 = (t^2-at+b)(t^2+at+c)`

`color(white)(t^4+10t^2+168t-1715) = t^4+(b+c-a^2)t^2+a(b-c)t+bc`

Then equating coefficients, we find:

`{ (b+c=a^2+10), (b-c=168/a), (bc=-1715) :}`

Then:

`(a^2)^2+20(a^2)+100 = (a^2+10)^2`

`color(white)((a^2)^2+20(a^2)+100) = (b+c)^2`

`color(white)((a^2)^2+20(a^2)+100) = (b-c)^2+4bc`

`color(white)((a^2)^2+20(a^2)+100) = 168^2/(a^2)+4(-1715)`

`color(white)((a^2)^2+20(a^2)+100) = 28224/(a^2)-6860`

Multiplying both ends by `a^2` and rearranging a little, this becomes a cubic in `a^2`...

`(a^2)^3+20(a^2)^2+6960(a^2)-28224 = 0`

Note that typically the cubic at this stage is much nastier to solve, but thankfully this cubic has one real positive rational root, namely:

`a^2=4`

So choose:

`a=2`

Then:

`b = 1/2(a^2+10+168/a) = 1/2(4+10+84) = 49`

`c = 1/2(a^2+10-168/a) = 1/2(4+10-84) = -35`

So:

`t^4+10t^2+168t-1715 = (t^2-2t+49)(t^2+2t-35)`

Then subtituting `t = (4x-3)`, we find:

`256(x^4–3x^3+4x^2-8)`

`= ((4x-3)^2-2(4x-3)+49)((4x-3)^2+2(4x-3)-35)`

`= (16x^2-24x+9-8x+6+49)(16x^2-24x+9+8x-6-35)`

`= (16x^2-32x+64)(16x^2-16x-32)`

`= 256(x^2-2x+4)(x^2-x-2)`

`= 256(x^2-2x+4)(x-2)(x+1)`

So:

`x^4-3x^3+4x^2-8 = (x^2-2x+4)(x-2)(x+1)`

Answer 3

It can be factored by seeking rational roots directly:

`x^4-3x^3+4x^2-8=(x+1)(x-2)(x^2-2x+4)`

Explanation:

The polynomial may be factored with little difficulty if you seek out its rational zeroes.

Rational zeroes of a polynomial having integer coefficients will have the form `\pm(p/q)` where `p` is a factor of the constant term (`-8`) and `q` is a factor of the leading coefficient (`1`). So the possible rational roots are:

`\pm 1,\pm 2, \pm 4, \pm 8`

A good way to test these roots is with synthetic division. Let's try this with `1` as a candidate root. Remember to include the `0` coefficient for `x^1` as you do the steps, otherwise you are testing the wrong polynomial!

Coefficients `color(blue)(1, -3, 4, 0, -8)`

Test root `color(red)(1)`

`color(blue)(1)×color(red)(1)=1`

`color(blue)(-3)+1=-2`

`(-2)×color(red)(1)=(-2)`

`color(blue)(4)+(-2)=2`

`2×color(red)(1)=2`

`color(blue)(0)+2=2`

`2×color(red)(1)=2`

`2+color(blue)(-8)=(-6)`

The final remainder term is `-6` not zero, so `x=1` fails to be a root and we do not have `x-1` as a factor.

But if we try `x=-1`:

`color(blue)(1)×color(red)(-1)=-1`

`color(blue)(-3)+(-1)=-4`

`(-4)×color(red)(-1)=4`

`color(blue)(4)+8=8`

`8×color(red)(-1)=-8`

`color(blue)(0)+(-8)=(-8)`

`(-8)×color(red)(-1)=8`

`8+color(blue)(-8)=0`

This time the remainder is `0` so we hit `x=-1` as a rational root. Therefore `x+1` is a factor. The leading coefficient `1` and the intermediate sums `-4,8,-8` form the polynomial `x^3-4x^2+8x-8`, which is the complementary factor. So:

`x^4-3x^3+4x^2-8=(x+1)(x^3-4x^2+8x-8)`

We are not done. The cubic factor `x^3-4x^2+8x-8` could have more rational roots. Again the possibilities are

`\pm 1,\pm 2, \pm 4, \pm 8`

But we know that `+1` doesn't work, and since the signs are strictly alternating (+-+-) there can't be any more negative roots (Descartes rule ofcsigns). The only ones we can still test are `+2,+4,+8`.

Try `+2` using synthetic division:

`color(blue)(1)×color(red)(2)=2`

`color(blue)(-4)+2=-2`

`(-2)×color(red)(2)=-4`

`color(blue)(8)+(-4)=4`

`4×color(red)(2)=8`

`color(blue)(-8)+8=0`

We get a remainder of zero and so

`x^3-4x^2+8x-8=(x-2)(x^2-2x+4)`

And then

`x^4-3x^3+4x^2-8=(x+1)(x^3-4x^2+8x-8)=(x+1)(x-2)(x^2-2x+4)`

We are finally left with the quadratic polynomial `x^2-2x+4`. Does this have any more rational roots? For a quadratic polynomial `ax^2+bx+c` we can tell if there are any real roots at all from the discriminant `b^2-4ac`:

`(-2)^2-(4×1×4)=-12`, negative thetefore no (moe) real roots. And obviously no more rational roots.

So the quadratic polynomial remains unfactored and this is the final answer:

`x^4-3x^3+4x^2-8=(x+1)(x-2)(x^2-2x+4)`