Can a triangle be formed with sides 18, 20, 2? If so, classify it by angles..

Mathematics

1 answer:

Nady [450]2 years ago

7 0

Answer:

No, it cannot make a triangle

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Find the polynomial of minimum degree, with real coefficients, zeros at

drek231 [11]

Answer:

$\huge\boxed{p(x)=4x^3-20x^2+4x+300}$

Step-by-step explanation:

$\text{If}\ x=4\pm3i\ \text{and}\ x=-3\ \text{are the zeros of a polynomial, then it has a form:}\\\\p(x)=\bigg(x-(4-3i)\bigg)\bigg(x-(4+3i)\bigg)\bigg(x-(-3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x-4+3i)(x-4-3i)(x+3)\bigg(r(x)\bigg)\\\\p(x)=\underbrace{\bigg((x-4)+3i\bigg)\bigg((x-4)-3i\bigg)}_{\text{use}\ (a+b)(a-b)=a^2-b^2}(x+3)\bigg(r(x)\bigg)\\\\p(x)=\bigg((x-4)^2-(3i)^2\bigg)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2$

$p(x)=(x^2-2(x)(4)+4^2-3^2i^2)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ i^2=-1\\\\p(x)=(x^2-8x+16-9(-1))(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+16+9)(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+25)(x+3)\bigg(r(x)\bigg)\qquad\text{use FOIL}:\ (a+b)(c+d)=ac+ad+bc+bd\\\\p(x)=\bigg((x^2)(x)+(x^2)(3)+(-8x)(x)+(-8x)(3)+(25)(x)+(25)(3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x^3+3x^2-8x^2-24x+25x+75)\bigg(r(x)\bigg)\qquad\text{combine like terms}\\\\p(x)=(x^3-5x^2+x+75)\bigg(r(x)\bigg)$

$\text{The y-intercept is at 300}.\\\\\text{For}\ w(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0\\\\\text{y-intercept is}\ a_0\\\\\text{Therefore for}\ p(x)=(x^3-5x^2+x+75)\bigg(r(x)\bigg)\\\\\text{y-intercet is}\ 75\bigg(r(x)\bigg)\\\\75\bigg(r(x)\bigg)=300\qquad\text{divide both sides by 75}\\\\r(x)=4\\\\\text{Finally:}\\\\p(x)=(x^3-5x^2+x+75)(4)\qquad\text{use the distributive property}\\\\p(x)=(x^3)(4)+(-5x^2)(4)+(x)(4)+(75)(4)\\\\p(x)=4x^3-20x^2+4x+300$