Question

1. $-105x^{2}+99x - 27=-25x^{3}$
2. $3x^{3}-13x^{2}+60x - 91=2x^{3}+5x$
3. $6x^{3}+16x=25x^{2}-15$
4. $31x^{2}+31x + 5=-5x^{3}$

Explanation:

Problem 7

Step1: Rearrange to standard form

$25x^3 - 105x^2 + 99x - 27 = 0$

Step2: Factor by grouping

Group terms: $(25x^3 - 105x^2) + (99x - 27) = 0$
Factor: $5x^2(5x - 21) + 9(11x - 3) = 0$
Test rational roots (Rational Root Theorem: possible roots $\pm1,\pm3,\pm9,\pm27,\pm\frac{1}{5},\pm\frac{3}{5},...$). Test $x=\frac{3}{5}$:
$25(\frac{3}{5})^3 -105(\frac{3}{5})^2 +99(\frac{3}{5}) -27 = 25(\frac{27}{125}) -105(\frac{9}{25}) + \frac{297}{5} -27 = \frac{27}{5} - \frac{189}{5} + \frac{297}{5} - \frac{135}{5} = 0$
Factor out $(5x-3)$: Use polynomial division or synthetic division
Divide $25x^3 -105x^2 +99x -27$ by $(5x-3)$:
$5x^2 - 18x +9$
Factor quadratic: $5x^2 -15x -3x +9 = 5x(x-3)-3(x-3)=(5x-3)(x-3)$
Final factored form: $(5x-3)^2(x-3)=0$

Step3: Solve for x

Set each factor to 0:
$5x-3=0 \implies x=\frac{3}{5}$; $x-3=0 \implies x=3$

Step1: Rearrange to standard form

$3x^3 -13x^2 +60x -91 -2x^3 -5x = 0$
$x^3 -13x^2 +55x -91 = 0$

Step2: Test rational roots

Possible roots: $\pm1,\pm7,\pm13,\pm91$. Test $x=7$:
$7^3 -13(7)^2 +55(7)-91 = 343 - 637 + 385 -91 = 0$

Step3: Factor polynomial

Divide $x^3 -13x^2 +55x -91$ by $(x-7)$:
$x^2 -6x +13$

Step4: Solve quadratic

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ for $x^2-6x+13=0$, $a=1,b=-6,c=13$
$x=\frac{6\pm\sqrt{36-52}}{2}=\frac{6\pm\sqrt{-16}}{2}=\frac{6\pm4i}{2}=3\pm2i$

Step1: Rearrange to standard form

$6x^3 -25x^2 +16x +15 = 0$

Step2: Test rational roots

Possible roots: $\pm1,\pm3,\pm5,\pm15,\pm\frac{1}{2},\pm\frac{3}{2},...$. Test $x=\frac{3}{2}$:
$6(\frac{3}{2})^3 -25(\frac{3}{2})^2 +16(\frac{3}{2}) +15 = 6(\frac{27}{8}) -25(\frac{9}{4}) +24 +15 = \frac{81}{4} - \frac{225}{4} + 39 = \frac{-144}{4}+39=-36+39=3
eq0$
Test $x=3$:
$6(27)-25(9)+16(3)+15=162-225+48+15=0$

Step3: Factor polynomial

Divide $6x^3 -25x^2 +16x +15$ by $(x-3)$:
$6x^2 -7x -5$
Factor quadratic: $6x^2 -10x +3x -5=2x(3x-5)+1(3x-5)=(2x+1)(3x-5)$

Step4: Solve for x

Set factors to 0:
$x-3=0 \implies x=3$; $2x+1=0 \implies x=-\frac{1}{2}$; $3x-5=0 \implies x=\frac{5}{3}$

Answer:

$x=\frac{3}{5}$ (multiplicity 2), $x=3$

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Problem 8