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)

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

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

$$24\mu=120$$

$$\mu=5$$