How do you solve $2 y^{3} - y^{2} - 2 + 1 \leq 0$ $2y^3-y^2-2+1<=0$ using a sign chart?

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How do you solve $2 y^{3} - y^{2} - 2 + 1 \leq 0$ $2y^3-y^2-2+1<=0$ using a sign chart?

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Answer 1 · 2017-09-30T19:22:31

The solution is $y \in (- \infty, - 1] \cup [\frac{1}{2}, 1]$ $y in (-oo,-1]uu[1/2,1]$

Explanation:

I assume that $- 2$ $-2$ means $- 2 y$ $-2y$

Before performing the sign chart, we need the roots of the polynomial

Let $f (y) = 2 y^{3} - y^{2} - 2 y + 1$ $f(y)=2y^3-y^2-2y+1$

$f (1) = 2 - 1 - 2 + 1 = 0$ $f(1)=2-1-2+1=0$

Therefore,

$(y - 1)$ $(y-1)$ is a factor of the polynomial

Therefore,

$2 y^{3} - y^{2} - 2 y + 1 = (y - 1) (a y^{2} + b y + c)$ $2y^3-y^2-2y+1=(y-1)(ay^2+by+c)$

$= a y^{3} + b y^{2} + c y - a y^{2} - b y - c$ $=ay^3+by^2+cy-ay^2-by-c$

Comparing the coefficients

$a = 2$ $a=2$

$b - a = - 1$ $b-a=-1$ , $\Rightarrow$ $=>$ , $b = a - 1 = 2 - 1 = 1$ $b=a-1=2-1=1$

$c - b = - 2$ $c-b=-2$ , $\Rightarrow$ $=>$ , $c = b - 2 = 1 - 2 = - 1$ $c=b-2=1-2=-1$

Therefore,

$2 y^{3} - y^{2} - 2 y + 1 = (y - 1) (2 y^{2} + y - 1) = (y - 1) (2 y - 1) (y + 1)$ $2y^3-y^2-2y+1=(y-1)(2y^2+y-1)=(y-1)(2y-1)(y+1)$

We can build the sign chart

$a a a a$ $color(white)(aaaa)$ $y$ $y$ $a a a a a$ $color(white)(aaaaa)$ $- \infty$ $-oo$ $a a a a a$ $color(white)(aaaaa)$ $- 1$ $-1$ $a a a a a a a$ $color(white)(aaaaaaa)$ $\frac{1}{2}$ $1/2$ $a a a a a a$ $color(white)(aaaaaa)$ $1$ $1$ $a a a a a a$ $color(white)(aaaaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $y + 1$ $y+1$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $0$ $0$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $2 y - 1$ $2y-1$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ #color(white)(aaaaa)-# $a$ $color(white)(a)$ $0$ $0$ $a a$ $color(white)(aa)$ $+$ $+$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $y - 1$ $y-1$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ #color(white)(aaaaa)-# $a a a$ $color(white)(aaa)$ #color(white)(a)-# $a a$ $color(white)(aa)$ $0$ $0$ $a a a$ $color(white)(aaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (y)$ $f(y)$ $a a a a a a a$ $color(white)(aaaaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $0$ $0$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a$ $color(white)(a)$ $0$ $0$ $a$ $color(white)(a)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $0$ $0$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (y) \leq 0$ $f(y)<=0$ when $y \in (- \infty, - 1] \cup [\frac{1}{2}, 1]$ $y in (-oo,-1]uu[1/2,1]$

graph{2x^3-x^2-2x+1 [-5.55, 5.55, -2.773, 2.776]}

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