Are All Zeros of This Polynomial Real?

[anemone]

Show that all zeros of the polynomial $P(x)=(x-1)(x-3)(x-5)(x-7)+x(x-2)(x-4)(x-6)$ are real.

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[anemone]

Congratulations to the following members for their correct solutions::)

1. magneto
2. mathlover
3. mathbalarka
4. RLBrown
5. kaliprasad
6. Olok
7. MarkFL
8. laura123

Solution from mathlover:

Spoiler

Using the fact , If P(a).P(b) is negative , then at least one real zero of polynomial P(x) lies in (a,b).
Given polynomial is of 4 degree and P(0).P(1), P(2).P(3), P(4).P(5) and P(6).P(7) are all negative.

Hence exactly one root lies in each of the following intervals:
(0,1) , (2,3) , (4,5) and (6,7)

The given proposition proves that all four roots of given polynomial are real.

Solution from mathbalarka:

Spoiler

It is clear that $P(x) = (x - 1)(x - 3)(x - 5)(x - 7) + x(x-2)(x-4)(x-6)$ is a quartic polynomial, and thus has a total of $4$ roots over $\Bbb C$. Note that

$$P(0) = (-1)(-3)(-5)(-7) + (0)(-2)(-4)(-6) = 105$$
$$P(1) = (0)(-3)(-5)(-7) + (1)(1-2)(1-4)(1-6) = (1)(-1)(-3)(-5) = -15$$
$$P(2) = (2-1)(2-3)(2-5)(2-7) + (2)(2-2)(2-4)(2-6) = (1)(-1)(-3)(-5) = -15$$
$$P(3) = (3-1)(3-3)(3-5)(3-7) + (3)(3-2)(3-4)(3-6) = (3)(1)(-1)(-3) = 9$$
$$P(4) = (4-1)(4-3)(4-5)(4-7) + (4)(4-2)(4-4)(4-6) = (3)(1)(-1)(-3) = 9$$
$$P(5) = (5-1)(5-3)(5-5)(5-7) + (5)(5-2)(5-4)(5-6) = (5)(3)(1)(-1) = -15$$
$$P(6) = (6-1)(6-3)(6-5)(6-7) + (6)(6-2)(6-4)(6-6) = (5)(3)(1)(-1) = -15$$
$$P(7) = (7-1)(7-3)(7-5)(7-7) + (7)(7-2)(7-4)(7-6) = (7)(5)(3)(1) = 105$$

As the polynomial $P(x)$ sign changes in the four intervals $[0, 1]$, $[2, 3]$ $[4, 5]$ and $[6, 7]$, the four roots are sitting inside each of the intervals, hence all of the roots are real.

Solution from RLBrown:

Spoiler

Making use of the symmetry of P(x) about x=7/2
Let u = x-7/2 then the symmetry of Q(u) about zero yeilds only even powers of u.
P(u+7/2) => Q(u) = 105/8 - 17u^2 + 2u^4

Q(u) can be rewritten as a quadratic R(v) by letting v=u^2
Q(u) => R(v) = 105/8 - 17 v + 2 v^2
the discriminant of R indicates that both roots v1 and v2 are real.
More importantly, both roots v1 and v2 are positive reals.

There are 4 real roots to Q(u), specifically $\pm \sqrt{\text{v1}}
\text{ and}
\pm \sqrt{\text{v2}}$
Therefore, all zeros of the polynomial P are real, by x = u + 7/2 = $\sqrt{v}$ + 7/2
$\blacksquare$

Solution from Olok:

Spoiler

$P(x) = (x-1)(x-3)(x-5)(x-7) + x(x-2)(x-4)(x-6) = 2x^4 - 28x^3 + 130x^2 - 224x + 105$
$P'(x) = 8x^3 - 84x^2 + 260x - 224 = 4(2x^3 - 21x^2 + 65x - 56)$
$P''(x) =4(6x^2 - 42x + 65)$

Since $P(x)$ is a polynomial the IVT can be applied as $P(x)$ is continuous for $ x \in \Bbb{R}$

Set $P'(x) = 0$ to find the extrema of $P(x)$ the function, $P'(x)$ does not seem factorable by regular methods; use the rational zero method and list the factors of $-56$ divided by factors of $2$, after a long and tedious task, one discovers by synthetic division, $x = 7/2$ is a rational zero, and it is possible to factor the function now (into simpler forms).

$P'(x) = 8(x - 7/2)(x^2 - 7x + 8)$ Now set $P'(x) = 0$ Use the discriminant for the quadratic & the quadratic formula, you find out solutions are $x = 7/2, \frac{7+\sqrt(17)}{2}, \frac{7-\sqrt(17)}{2}$ you recognize, $\frac{7-\sqrt(17)}{2} < 7/2 < \frac{7+\sqrt(17)}{2}$

$\frac{7-\sqrt(17)}{2} \approx (7-4)/2 = 3/2$ and $\frac{7+\sqrt(17)}{2} \approx (7+4)/2 = 11/2 = 5.5$

$P''(\frac{7+\sqrt(17)}{2}) > 0$ and $P''(\frac{7-\sqrt(17)}{2}) > 0$ and $P''(7/2) < 0$

For those $x$ for which $P''(x) < 0$ those $x$'s are maximum and for those $x$'s which $P''(x) > 0$ those $x$'s are minimums (concave up).

First we substitute the value of the critical numbers into $P(x)$ to identify their values (signs).

$P(\frac{7-\sqrt(17)}{2}) \approx 2(1.5)^4 - 28(1.5)^3 + 130(1.5)^2 - 224(1.5) + 105 = -22< 0$ Using $\sqrt(17) \approx 4$

$P(7/2) = P(3.5) \approx 13 > 0$

$P(\frac{7+\sqrt(17)}{2}) \approx 2(5.5)^4 - 28(5.5)^3 + 130(5.5)^2 - 224(5.5) + 105 = -22 < 0$ Using $\sqrt(17) \approx 4$

Thus, between $(\frac{7-\sqrt(17)}{2}, 7/2)$ the IVT is satisfied, and there exists a point $A$ such that $P(A) = 0$
Between $(7/2, \frac{7+\sqrt(17)}{2})$ the IVT is satisfied, and there exists a point $B$ such that $P(B) = 0$

We need to consider four $x$ intervals

$(-\infty, \frac{7-\sqrt(17)}{2})$ and $(\frac{7-\sqrt(17)}{2}, 7/2)$ and $(7/2, \frac{7+\sqrt(17)}{2})$ and $(\frac{7+\sqrt(17)}{2}, +\infty)$

Consider first, $(-\infty, \frac{7-\sqrt(17)}{2})$ test a point $x = 0$ which gives $P'(0) = 4(-56) < 0$ $P(x)$ is decreasing on this. This will happen from $(-\infty, \frac{7-\sqrt(17)}{2})$ so it is advisable to test a point extremely small for more sign of $P(x)$ accuracy. But $x = 0$ works here.

$P(0) = 105 > 0$ and the first relative minimum point is $x = \frac{7-\sqrt(17)}{2}$ and $P(\frac{7-\sqrt(17)}{2}) \approx -22 < 0$

Therefore, by the IVT, there exists a point $C$ in $(-\infty, \frac{7-\sqrt(17)}{2})$ such that $P(C) = 0$

Consider last, $(\frac{7+\sqrt(17)}{2}, +\infty)$ and consider that $\frac{7+\sqrt(17)}{2} \approx (7+4)/2 = 5.5$

Test for example $x = 10$
$P'(10) > 0$ therefore, $P(x)$ is increasing on the interval. Since the last relative extrema is $x = \frac{7+\sqrt(17)}{2}$, the increasing behavior continues forever, a point should be checked in $P(x)$ now. For example, $x = 10$

$P(10) = 2865 > 0$

Recall that $P(\frac{7+\sqrt(17)}{2}) \approx -22 < 0$

Therefore, the IVT suggests there exists a point $D$ in $(\frac{7+\sqrt(17)}{2}, +\infty)$ such that $P(D) = 0$. This completes the proof.

We have proved by the IVT, that there exist four points,

$A \in (\frac{7-\sqrt(17)}{2}, 7/2)$ such that $P(A) = 0$
$B \in (7/2, \frac{7+\sqrt(17)}{2})$ such that $P(B) = 0$
$C \in (-\infty, \frac{7-\sqrt(17)}{2})$ such that $P(C) = 0$
$D \in (\frac{7+\sqrt(17)}{2}, +\infty)$ such that $P(D) = 0$

This proves that all zeros are real because $P(x)$ is a fourth degree polynomial; a maximum of four zeros is applicable, which we have proved by the IVT to be within the real-coordinate system.