Answer to Question #145498 in Algebra for Rolando

Question

1. 3x² + 2x – 1=0
Ans.

2. x³-27
Ans.

State the possible rational zeros for each polynomial. Then find all rational zeros.
3. P(x) x³ +4x² + 5x + 2
Possible zeros:
Rational zeros:

4. 3x³ + 11x² + 5x – 3
Possible zeros:
Rational zeros:

Answer 1

1.

$3x² + 2x – 1=0$

Here, we can apply quadratic formula

$x = \frac{−b± \sqrt{b^2−4ac} }{2a}$

Where a = 3, b = 2, c = -1

Plugging these into the equation, we have;

$x = \frac{−2± \sqrt{2^2−(4×3×-1)} }{2×-1}$

$x = \frac{−2±√16}{6}$

$x= \frac{1}{3}$ or $x = -1$

2.

$x^3 - 27 =0$ -27 crosses the other side

$\implies x^3 = 27$ take the cube root of both sides

$\implies \sqrt[3]{x^3} = \sqrt[3]{27}$

$x = 3$

3.

$\mathrm{For\:a\:polynomial\:equation\:with\:integer\:coefficients:\quad }a_nx^n+a_{n-1}x^{n-1}+\ldots +a_0$

$\mathrm{If\:}a_0\mathrm{\:and\:}a_n\mathrm{\:are\:integers,\:then\:if\:there\:is\:a\:rational\:solution}$

$it\:could\:be\:found\:by\:checking\:all\:the\:numbers\:produced$

$\:for\:\pm \frac{dividers\:of\:a_0}{dividers\:of\:a_n}$

Therefore, given the polynomial x³ +4x² + 5x + 2, we have;

$a_0=2, and\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2, and\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

So, $\mathrm{The\:possible\:zeros\:numbers\:are\:\:}\quad \pm \frac{1,\:2}{1}$ or $\quad \pm \frac{1}{1}$ , $\quad \pm \frac{2}{1}$

And the rational zeros are obtained when these are replaced in the polynomial $x^3+4x^2+5x+2$ , we have $x = -1, and , x=-2$

4.

$3x³ + 11x² + 5x – 3$

The above procedure applies to this polynomial where;

$a_0=3, and\:\quad a_n=3$

And $\mathrm{the\:dividers\:of\:}a_0:\quad 1,\:3,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:3$

Hence the possible zeros are $\pm \frac{1,\:3}{1,\:3}$

And the rational zeros are, when plugged back into the polynomial

$3x³ + 11x² + 5x – 3$ , we have; $x=-1,\:x=-3,\:x=\frac{1}{3}$