kukuhushlq3p

2023-02-18

How to find all the real solutions of the polynomial equation $2{y}^{4}+7{y}^{3}-26{y}^{2}+23y-6=0$?

Adrienne Clark

Beginner2023-02-19Added 9 answers

$2{y}^{4}+7{y}^{3}-26{y}^{2}+23y-6$

Keep in mind that $0$ is the total of the coefficients. Which is:

$2+7-26+23-6=0$

So $y=1$ is a zero and $(y-1)$ a factor:

$2{y}^{4}+7{y}^{3}-26{y}^{2}+23y-6=(y-1)(2{y}^{3}+9{y}^{2}-17y+6)$

The sum of the coefficients of the remaining cubic factor also sum to $0$. That is:

$2+9-17+6=0$

So $y=1$ is a zero again and $(y-1)$ a factor:

$2{y}^{3}+9{y}^{2}-17y+6=(y-1)(2{y}^{2}+11y-6)$

We can factor the remaining quadratic using an AC method:

Find a pair of factors of $AC=2\cdot 6=12$ which differ by $B=11$

The pair $12,1$ works.

Split the middle term using that pair, then factor by grouping:

$2{y}^{2}+11y-6$

$=2{y}^{2}+12y-y-6$

$=(2{y}^{2}+12y)-(y+6)$

$=2y(y+6)-1(y+6)$

$=(2y-1)(y+6)$

So the remaining zeros are $y=\frac{1}{2}$ and $y=-6$

Keep in mind that $0$ is the total of the coefficients. Which is:

$2+7-26+23-6=0$

So $y=1$ is a zero and $(y-1)$ a factor:

$2{y}^{4}+7{y}^{3}-26{y}^{2}+23y-6=(y-1)(2{y}^{3}+9{y}^{2}-17y+6)$

The sum of the coefficients of the remaining cubic factor also sum to $0$. That is:

$2+9-17+6=0$

So $y=1$ is a zero again and $(y-1)$ a factor:

$2{y}^{3}+9{y}^{2}-17y+6=(y-1)(2{y}^{2}+11y-6)$

We can factor the remaining quadratic using an AC method:

Find a pair of factors of $AC=2\cdot 6=12$ which differ by $B=11$

The pair $12,1$ works.

Split the middle term using that pair, then factor by grouping:

$2{y}^{2}+11y-6$

$=2{y}^{2}+12y-y-6$

$=(2{y}^{2}+12y)-(y+6)$

$=2y(y+6)-1(y+6)$

$=(2y-1)(y+6)$

So the remaining zeros are $y=\frac{1}{2}$ and $y=-6$