Answer to Question #158701 in Algebra for moon

Question

Answer the following problems.

2.) Determine the sum pf the real zeros of P(x) = 3x⁴ + 14x³ + 10x² - 14x + 3.

3.) Find μ so that x - 3 is a factor of  μx³ - 13x³ - μx - 3.

Answer 1

2)

$x=-3$

$3(-3)^4+14(-3)^3+10(-3)^2-14(-3)+3=0$

$3x^4+14x^3+10x^2-14x+3=$

$=3x^3(x+3)+5x^2(x+3)-5x(x+3)+x+3=$

$=3(x+3)(x^3+\dfrac{5}{3}x^2-\dfrac{5}{3}x+\dfrac{1}{3})$

$=3(x+3)(x^3-\dfrac{1}{3}x^2+2x^2-\dfrac{2}{3}x-x+\dfrac{1}{3})$

$=3(x+3)(x-\dfrac{1}{3})(x^2+2x-1)$

$=3(x+3)(x-\dfrac{1}{3})(x+1-\sqrt{2})(x+1+\sqrt{2})$

$x_1+x_2+x_3+x_4=-3+\dfrac{1}{3}-2=-\dfrac{14}{3}$

Vieta's formulas

$x_1+x_2+x_3+x_4=-\dfrac{a_3}{a_4}=-\dfrac{14}{3}$

$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4$

$=\dfrac{a_2}{a_4}=\dfrac{10}{3}$

$x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4$

$=-\dfrac{a_1}{a_4}=\dfrac{14}{3}$

$x_1x_2x_3x_4=\dfrac{a_0}{a_4}=1$

3)

$P(x)=\mu x^3-13x^2-\mu x-3$

$P(3)=\mu (3)^3-13(3)^2-\mu (3)-3=0$

$24\mu=120$

$\mu=5$