Polynomial Equations of Higher Degree: Problems with Solutions

Solution:
Since [tex]x+2=x-(-2)[/tex], we apply Horner's scheme with [tex]\alpha=-2[/tex] to the coefficients [tex]6, -1, 11, -2, -2[/tex]. Each new number equals [tex]\alpha[/tex] times the previous result plus the next coefficient:
[tex]6, \quad (-2)\times 6+(-1)=-13, \quad (-2)\times (-13)+11=37, \quad (-2)\times 37+(-2)=-76, \quad (-2)\times (-76)+(-2)=150[/tex]
The first four numbers are the coefficients of the quotient, and the last one is the remainder:
[tex]f(x)=(x+2)\left(6x^{3}-13x^{2}+37x-76\right)+150[/tex]
Check with the remainder theorem: [tex]f(-2)=96+8+44+4-2=150[/tex]. The remainder is [tex]150[/tex].

Solution:
By the remainder theorem (Bézout's theorem), the remainder of the division of [tex]P(x)[/tex] by [tex]x-3[/tex] equals [tex]P(3)[/tex]:
[tex]P(3)=81-81+9-3k+5=14-3k[/tex]
The condition of the problem gives
[tex]14-3k=2\Rightarrow k=4[/tex]
Check with Horner's scheme for [tex]k=4[/tex]: dividing the coefficients [tex]1, -3, 1, -4, 5[/tex] by [tex]x-3[/tex] gives the quotient coefficients [tex]1, 0, 1, -1[/tex] and the remainder [tex](-1)\times 3+5=2[/tex].

Solution:
Multiplying by [tex]-1[/tex], we get a biquadratic equation:
[tex]x^4-4x^2-32=0[/tex]
Put [tex]t=x^2\ge 0[/tex]:
[tex]t^2-4t-32=0\Rightarrow t_1=8,\ t_2=-4[/tex]
The value [tex]t_2=-4[/tex] is rejected, because [tex]x^2[/tex] cannot be negative. From [tex]x^2=8[/tex] we get
[tex]x_{1,2}=\pm\sqrt{8}=\pm 2\sqrt{2}[/tex]

Solution:
The coefficients [tex]1, 1, 3, 3, 1, 1[/tex] are symmetric, so this is a symmetric (reciprocal) equation of odd degree and [tex]x=-1[/tex] is a root. Grouping the terms in pairs, we get
[tex]x^4(x+1)+3x^2(x+1)+(x+1)=0\iff (x+1)\left(x^4+3x^2+1\right)=0[/tex]
For every real [tex]x[/tex] we have [tex]x^4+3x^2+1\ge 1\gt 0[/tex], so the second factor has no real zeros. The only real root is [tex]x=-1[/tex].

Solution:
Let [tex]P(x)=x^4-4x^3-19x^2+106x-120[/tex]. All coefficients are integers and the leading coefficient is [tex]1[/tex], so every rational root is an integer divisor of [tex]120[/tex]. We check:
[tex]P(1)=1-4-19+106-120=-36\ne 0,\quad P(2)=16-32-76+212-120=0[/tex]
Horner's scheme with [tex]2[/tex] on the coefficients [tex]1, -4, -19, 106, -120[/tex] gives [tex]1, -2, -23, 60[/tex] and remainder [tex]0[/tex], so
[tex]P(x)=(x-2)\left(x^3-2x^2-23x+60\right)[/tex]
For the cubic factor, [tex]x=3[/tex] is a root ([tex]27-18-69+60=0[/tex]), and Horner's scheme with [tex]3[/tex] gives [tex]1, 1, -20[/tex]. Hence
[tex]x^3-2x^2-23x+60=(x-3)\left(x^2+x-20\right)=(x-3)(x-4)(x+5)[/tex]
Thus [tex]P(x)=(x-2)(x-3)(x-4)(x+5)[/tex] and the roots of the equation are
[tex]x\in\left\{-5, 2, 3, 4\right\}[/tex]

Solution:
For [tex]x\le 0[/tex] each of the terms [tex]6x^{4}, -23x^{3}, 37x^{2}, -28x[/tex] is non-negative and the free term is [tex]6\gt 0[/tex], so the left side is positive; hence every real root is positive. A rational root [tex]\frac{p}{q}[/tex] (in lowest terms) has [tex]p[/tex] dividing [tex]6[/tex] and [tex]q[/tex] dividing [tex]6[/tex], so the positive candidates are
[tex]1, 2, 3, 6, \frac{1}{2}, \frac{3}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{6}[/tex]
Horner's scheme with [tex]\frac{3}{2}[/tex]:
[tex]6, \quad 6\times \frac{3}{2}-23=-14, \quad (-14)\times \frac{3}{2}+37=16, \quad 16\times \frac{3}{2}-28=-4, \quad (-4)\times \frac{3}{2}+6=0[/tex]
So [tex]\frac{3}{2}[/tex] is a root and the quotient is
[tex]6x^{3}-14x^{2}+16x-4=2\left(3x^{3}-7x^{2}+8x-2\right)[/tex]
Horner's scheme for [tex]3x^{3}-7x^{2}+8x-2[/tex] with [tex]\frac{1}{3}[/tex]:
[tex]3, \quad 3\times \frac{1}{3}-7=-6, \quad (-6)\times \frac{1}{3}+8=6, \quad 6\times \frac{1}{3}-2=0[/tex]
So [tex]\frac{1}{3}[/tex] is a root and the quotient is [tex]3x^{2}-6x+6=3\left(x^{2}-2x+2\right)[/tex]. Thus
[tex]6x^{4}-23x^{3}+37x^{2}-28x+6=(2x-3)(3x-1)\left(x^{2}-2x+2\right)[/tex]
The last factor [tex]x^{2}-2x+2=(x-1)^{2}+1[/tex] is always positive, so it gives no real roots. The real roots are
[tex]x\in\left\{\frac{1}{3}, \frac{3}{2}\right\}[/tex]

Solution:
The number [tex]x=-1[/tex] is a multiple root exactly when the polynomial is divisible by [tex](x+1)^{2}[/tex], i.e. when [tex]-1[/tex] is a root of the polynomial and also a root of the quotient after division by [tex]x+1[/tex]. Horner's scheme with [tex]-1[/tex] on the coefficients [tex]3, 4, -6, b, a[/tex]:
[tex]3, \quad -3+4=1, \quad -1-6=-7, \quad 7+b, \quad -(7+b)+a=a-b-7[/tex]
The first condition is [tex]a-b-7=0[/tex], and the quotient is [tex]3x^{3}+x^{2}-7x+(b+7)[/tex]. Horner's scheme with [tex]-1[/tex] on [tex]3, 1, -7, b+7[/tex]:
[tex]3, \quad -3+1=-2, \quad 2-7=-5, \quad 5+b+7=b+12[/tex]
The second condition is [tex]b+12=0[/tex]. Hence
[tex]b=-12,\quad a=b+7=-5[/tex]
Check: [tex]3x^{4}+4x^{3}-6x^{2}-12x-5=(x+1)^{3}(3x-5)[/tex], so for these values [tex]x=-1[/tex] is even a triple root.

Solution:
Multiply the first factor by the fourth and the second by the third:
[tex]x(x-3)=x^2-3x,\quad (x-1)(x-2)=x^2-3x+2[/tex]
With [tex]t=x^2-3x[/tex] the equation becomes
[tex]t(t+2)=-1\iff t^2+2t+1=0\iff (t+1)^2=0\iff t=-1[/tex]
Returning to [tex]x[/tex]:
[tex]x^2-3x+1=0,\quad D=9-4=5,\quad x_{1,2}=\frac{3\pm\sqrt{5}}{2}[/tex]
In fact the equation is [tex]\left(x^2-3x+1\right)^2=0[/tex], so both roots are double; this is why a search for rational roots with Horner's scheme fails here.