Respuesta :

gmany

a, b and c are the zeros of a polynomial: w(x) = (x - a)(x - b)(x - c).


[tex]24.\\w(x)=(x-6)[x-(-5-2i)][x-(-5+2i)]\\\\w(x)=(x-6)(x+5+2i)(x+5-2i)\\\\w(x)=(x-6)[(x+5)+2i][(x+5)-2i]\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\w(x)=(x-6)[(x+5)^2-(2i)^2]\\\\\text{use}\ (a+b)^2=a^2+2ab+b^2\ \text{and}\ (ab)^n=a^nb^n\\\\w(x)=(x-6)(x^2+2(x)(5)+5^2-2^2i^2)\\\\w(x)=(x-6)(x^2+10x+25-4(-1))\\\\w(x)=(x-6)(x^2+10x+25+4)[/tex]

[tex]w(x)=(x-6)(x^2+10x+29)\\\\\text{use distributive property}\ a(b+c)=ab+ac\\\\w(x)=(x)(x^2)+(x)(10x)+(x)(29)+(-6)(x^2)+(-6)(10x)+(-6)(29)\\\\w(x)=x^3+10x^2+29x-6x^2-60x-174\\\\\text{combine like terms}\\\\w(x)=x^3+(10x^2-6x^2)+(29x-60x)-174\\\\\boxed{w(x)=x^3+4x^2-31x-174}[/tex]


[tex]25.\\w(x)=(x-i)(x-(-i))(x-6i)(x-(-6i))\\\\w(x)=(x-i)(x+i)(x-6i)(x+6i)\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\w(x)=(x^2-i^2)(x^2-(6i)^2)\\\\\text{use}\ (ab)^n=a^nb^n\ \text{and}\ i^2=-1\\\\w(x)=(x^2-(-1))(x^2-6^2i^2)\\\\w(x)=(x^2+1)(x^2-36(-1))\\\\w(x)=(x^2+1)(x^2+36)\\\\\text{use distributive property}\ a(b+c)=ab+ac\\\\w(x)=(x^2)(x^2)+(x^2)(36)+(1)(x^2)+(1)(36)\\\\w(x)=x^4+36x^2+x^2+36\\\\\boxed{w(x)=x^4+37x^2+36}[/tex]