Math help- polynomials (2 Viewers)

CM_Tutor · May 28, 2020

Another approach is to look at the polynomial you seek to solve / factorise and separate the terms that can take on a odd value (restricting yourself to integers). Looking at the polynomial we have, every term will be even when x is an even integer, but when x is odd, there are only two terms that have an odd value, x⁴ and 19x². So, we can say:

$\begin{align*} \text{Set } P(x) = x^4 - 6x^3 - 19x^2 + 84x + 180 &= 0 \\ x^4 - 19x^2 &= 6x^3 - 84x - 180 \\ x^2(x^2 - 19) &= 6(x^3 - 14x - 30) \\ \cfrac{x^2(x^2 - 19)}{6} &= x^3 - 14x - 30 \end{align*}$

I can see that any odd integer must give a value of $x^2(x^2 - 19)$ that is a multiple of 6, which allows for a quick search for odd integer solutions.

Case: $x = \pm1$

$\text{LHS} = \cfrac{x^2(x^2 - 19)}{6} = \cfrac{(\pm1)^2\big((\pm1)^2 - 19\big)}{6} = \cfrac{1 \times -18}{6} = -3$
$\text{RHS} = x(x^2 - 14) - 30 = x\big((\pm1)^2 - 14\big) - 30 = -13x - 30 < -3 \text{ for } x = \pm1$
So, neither $x = 1$ nor $x = -1$ is a solution.

Case: $x = \pm3$

$\text{LHS} = \cfrac{x^2(x^2 - 19)}{6} = \cfrac{(\pm3)^2\big((\pm3)^2 - 19\big)}{6} = \cfrac{9 \times -10}{6} = -15$
$\text{RHS} = x(x^2 - 14) - 30 < -30 \text{ if } x = \pm3 > 0$
$\text{So, we need only test } x = -3 \text{: RHS } = -3 \times -5 - 30 = -15 = \text{ LHS}$
So, we have one solution, $x = -3$ .

Case: $x = \pm5$

$\text{LHS} = \cfrac{(\pm5)^2\big((\pm5)^2 - 19\big)}{6} = \cfrac{25 \times 6}{6} = 25$
$\text{RHS} = x(x^2 - 14) - 30 < 0 \text{ if } x = \pm5 < 0$
$\text{So, we need only test } x = +5 \text{: RHS } = 5 \times 11 - 30 = 25 = \text{ LHS}$
So, we have a second solution, $x = 5$ .

Having found two factors, we can divide $P(x)$ by $(x - 5)(x + 3) = x^2 - 2x - 15$ to find the second quadratic factor, which will be $x^2 - 4x - 12$ , but other approaches are faster.

We know that we can take the rots to be $\alpha$ , $\beta$ , $\gamma$ , and $\delta$ . We know two of these, so set $\gamma = -3$ and $\delta = 5$ . We also know (from the constant being positive) that there are an even number of positive roots, and so one of the two roots we are yet to find is positive and the other is negative.

$\text{Summing the roots:} \quad \alpha + \beta + -3 + 5 = \cfrac{-b}{a} = \cfrac{-(-6)}{1} \implies \alpha + \beta = 4$
$\text{Product of roots:} \quad \alpha \times \beta \times -3 \times 5 = \cfrac{e}{a} = \cfrac{180}{1} \implies \alpha\beta = \cfrac{180}{-15} = -12$

It immediately follows that the quadratic with roots $\alpha$ and $\beta$ is $x^2 - (\alpha + \beta)x + \alpha\beta = x^2 - 4x - 12 = (x - 6)(x + 2)$ which is also what the long division would find.

Thus, we have:
$P(x) = x^4 - 6x^3 - 19x^2 + 84x - 180 = (x + 2)(x + 3)(x - 5)(x - 6)$
and thus the solutions of $P(x) = 0$ are $x= -3, -2, 5, \text{ or } 6$ .

B1andB2 · May 28, 2020

CM_Tutor said:
Another approach is to look at the polynomial you seek to solve / factorise and separate the terms that can take on a odd value (restricting yourself to integers). Looking at the polynomial we have, every term will be even when x is an even integer, but when x is odd, there are only two terms that have an odd value, x⁴ and 19x². So, we can say:

$\begin{align*} \text{Set } P(x) = x^4 - 6x^3 - 19x^2 + 84x + 180 &= 0 \\ x^4 - 19x^2 &= 6x^3 - 84x - 180 \\ x^2(x^2 - 19) &= 6(x^3 - 14x - 30) \\ \cfrac{x^2(x^2 - 19)}{6} &= x^3 - 14x - 30 \end{align*}$

I can see that any odd integer must give a value of $x^2(x^2 - 19)$ that is a multiple of 6, which allows for a quick search for odd integer solutions.

Case: $x = \pm1$

$\text{LHS} = \cfrac{x^2(x^2 - 19)}{6} = \cfrac{(\pm1)^2\big((\pm1)^2 - 19\big)}{6} = \cfrac{1 \times -18}{6} = -3$
$\text{RHS} = x(x^2 - 14) - 30 = x\big((\pm1)^2 - 14\big) - 30 = -13x - 30 < -3 \text{ for } x = \pm1$
So, neither $x = 1$ nor $x = -1$ is a solution.

Case: $x = \pm3$

$\text{LHS} = \cfrac{x^2(x^2 - 19)}{6} = \cfrac{(\pm3)^2\big((\pm3)^2 - 19\big)}{6} = \cfrac{9 \times -10}{6} = -15$
$\text{RHS} = x(x^2 - 14) - 30 < -30 \text{ if } x = \pm3 > 0$
$\text{So, we need only test } x = -3 \text{: RHS } = -3 \times -5 - 30 = -15 = \text{ LHS}$
So, we have one solution, $x = -3$ .

Case: $x = \pm5$

$\text{LHS} = \cfrac{(\pm5)^2\big((\pm5)^2 - 19\big)}{6} = \cfrac{25 \times 6}{6} = 25$
$\text{RHS} = x(x^2 - 14) - 30 < 0 \text{ if } x = \pm5 < 0$
$\text{So, we need only test } x = +5 \text{: RHS } = 5 \times 11 - 30 = 25 = \text{ LHS}$
So, we have a second solution, $x = 5$ .

Having found two factors, we can divide $P(x)$ by $(x - 5)(x + 3) = x^2 - 2x - 15$ to find the second quadratic factor, which will be $x^2 - 4x - 12$ , but other approaches are faster.

We know that we can take the rots to be $\alpha$ , $\beta$ , $\gamma$ , and $\delta$ . We know two of these, so set $\gamma = -3$ and $\delta = 5$ . We also know (from the constant being positive) that there are an even number of positive roots, and so one of the two roots we are yet to find is positive and the other is negative.

$\text{Summing the roots:} \quad \alpha + \beta + -3 + 5 = \cfrac{-b}{a} = \cfrac{-(-6)}{1} \implies \alpha + \beta = 4$
$\text{Product of roots:} \quad \alpha \times \beta \times -3 \times 5 = \cfrac{e}{a} = \cfrac{180}{1} \implies \alpha\beta = \cfrac{180}{-15} = -12$

It immediately follows that the quadratic with roots $\alpha$ and $\beta$ is $x^2 - (\alpha + \beta)x + \alpha\beta = x^2 - 4x - 12 = (x - 6)(x + 2)$ which is also what the long division would find.

Thus, we have:
$P(x) = x^4 - 6x^3 - 19x^2 + 84x - 180 = (x + 2)(x + 3)(x - 5)(x - 6)$
and thus the solutions of $P(x) = 0$ are $x= -3, -2, 5, \text{ or } 6$ .

wow, thanks!

CM_Tutor · May 28, 2020

B1andB2 said:
wow, thanks!

No problem... another possibility in this kind of a situation is to see if a rearrangement can give you a graphical way to limit the possible domain of the solutions. Here, we have

$\cfrac{x^2(x^2 - 19)}{6} = x^3 - 14x - 30$

The graph of $y = \text{ LHS}$ is easy to sketch - it is an even quartic, has a double root (and thus a stationary point) at the origin, and its other roots are at $x = \pm \sqrt{19} = \pm 4.35...$ . Since $y \to +\infty$ as $x \to +\infty$ , it is clear that the origin will be a maximum turning point and there will be minima at some value $x = \pm a$ where $a \in (0, \sqrt{19})$ .

$\text{Since } \cfrac{d\text{LHS}}{dx} = \cfrac{4x^3 - 38x}{6} = \cfrac{x(2x^2 - 19)}{3}$
it is clear that the other stationary points (which we know are minima) occur when
$x^2 = \frac{19}{2} \implies x \approx \pm3.1$
$\text{where LHS } = \cfrac{\frac{19}{2}\big(\frac{19}{2} - 19\big)}{6} = \cfrac{\frac{19^2}{2}\big(\frac{1}{2} - 1\big)}{6} = \cfrac{-19^2}{2^2} \times \cfrac{1}{6} = -15\cfrac{1}{24}$
ie., they are at roughly $(\pm 3.1, -15)$

The graph of $y = \text{ RHS}$ is not so easily sketched, but it will match a general cubic with two stationary points. It has a y-intercept at (0, -30) - so more than twice as far below the x-axis as the minima for $y = \text{ LHS}$ - and its stationary points are accesible via calculus:
$\cfrac{d\text{RHS}}{dx} = 3x^2 - 14$ which means its stationary points are at $x^2 = \frac{14}{3} \implies x \approx \pm 2.2$
$\text{Since RHS } = x(x^2 - 14) - 30 = x(\frac{14}{3} - 14) - 30 = \cfrac{-28x}{3} - 30$ ,
the stationary points are near (2.2, -50) and (-2.2, -9.5).

The sketch can be done quickly if you consider the features of the first graph:

Take $x = 4$ (near but prior to the root of LHS at 4.35...):

$\text{LHS } = \cfrac{4^2(4^2 - 19)}{6} = -8$
$\text{RHS } = x(x^2 - 14) - 30 = 4(4^2 - 14) - 30 = -22$
So, LHS is well above RHS

Take $x = 5$ (just after the root of LHS):

$\text{LHS } = \cfrac{5^2(5^2 - 19)}{6} = 25$
$\text{RHS } = x(x^2 - 14) - 30 = 5(5^2 - 14) - 30 = 25$
So, LHS = RHS

Take $x = 3$ (near TP of LHS)

$\text{LHS } = \cfrac{3^2(3^2 - 19)}{6} = -15$
$\text{RHS } = x(x^2 - 14) - 30 = 3(3^2 - 14) - 30 = -45$
So, LHS is well above RHS

We know between x = 0 and x = 3 that RHS falls from y = -30 to y = -50 and then rises to y = -45, so is below LHS at all times

Take $x = -2$ (near TP of RHS)

$\text{LHS } = \cfrac{(-2)^2\big((-2)^2 - 19\big)}{6} = \cfrac{4 \times -15}{6} = -10$
$\text{RHS } = x(x^2 - 14) - 30 = -2\big((-2)^2 - 14\big) - 30 = -10$
So, LHS = RHS

Take $x = -3$ (near TP of LHS)

$\text{LHS } = \cfrac{(-3)^2\big((-3)^2 - 19\big)}{6} = \cfrac{9 \times -10}{6} = -15$
$\text{RHS } = x(x^2 - 14) - 30 = -3\big((-3)^2 - 14\big) - 30 = -15$
So, LHS = RHS

There will be no roots for $x < -3$ as RHS is increasing towards -15 and LHS is decreasing towards -15 where they meet (at $x = -3$ )

To check, take $x = -5$ (near root of LHS):

$\text{LHS } = \cfrac{(-5)^2\big((-5)^2 - 19\big)}{6} = \cfrac{25 \times 6}{6} = 25$
$\text{RHS } = x(x^2 - 14) - 30 = -5\big((-5)^2 - 14\big) - 30 = -85$
So, LHS is well above RHS

By chance, sketching LHS and RHS gives three of the roots, but the idea is looking near intercepts / stationary points, one can see regions within which solutions are likely:

There will be no roots in $x < -3$ , but had $x = -3$ not been a root, we could still have been certain of no roots in $x < -4.35...$
The positions of the stationary points near $x = -2.2$ and $x = -3.1$ suggests the presence of two roots in the region $-4 < x < 0$
There will be no roots in $0 < x < 4$ with the LHS clearly far above the RHS throughout this region
Beyond $x = 4$ , there must been either zero roots or an even number of roots because the LHS (a quartic) must end up above the RHS (a cubic)... unless the curves happen to meet at a point where they have a common tangent. Given we know there is a root at $x = 5$ and have found the ones at $x = -2$ and $x = -3$ , the place to seek the last one is in the region $x > 4$ . Our first guess would be $x = 6$ , which is correct.

Skecthing may not be that quick (depending on the graphs) but it does tell us the most likely candidates for integer roots here are $x = -1, -2, -3,$ and $x = 5, 6$ - a list containing all 4 of the sought roots.

Accurate · May 29, 2020

Will using synthetic division be 'proper' working out in an exam?

Velocifire · May 29, 2020

Yeah, i used it got 3/3 or 4/4

had to use g/c to find one factor then go from there

unless, they specifically ask for a certain way to do it of course, if not, it's fair game

Math help- polynomials (2 Viewers)

CM_Tutor

Moderator

B1andB2

oui oui baguette

CM_Tutor

Moderator

Accurate

clix

Velocifire

Critical Hit

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