kukuhushlq3p

2023-02-18

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

$2{y}^{4}+7{y}^{3}-26{y}^{2}+23y-6$
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 $\left(y-1\right)$ a factor:
$2{y}^{4}+7{y}^{3}-26{y}^{2}+23y-6=\left(y-1\right)\left(2{y}^{3}+9{y}^{2}-17y+6\right)$
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 $\left(y-1\right)$ a factor:
$2{y}^{3}+9{y}^{2}-17y+6=\left(y-1\right)\left(2{y}^{2}+11y-6\right)$
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:
$2{y}^{2}+11y-6$
$=2{y}^{2}+12y-y-6$
$=\left(2{y}^{2}+12y\right)-\left(y+6\right)$
$=2y\left(y+6\right)-1\left(y+6\right)$
$=\left(2y-1\right)\left(y+6\right)$
So the remaining zeros are $y=\frac{1}{2}$ and $y=-6$

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