Question #ec668

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Answer 1 · 2017-07-31T00:46:35

$P = - 3$ $P=-3$

Let $y^{2} - P y - 10$ $y^2-Py-10$ be $f (y)$ $f(y)$ .

Since $y - 2$ $y-2$ is a factor of $f (y)$ $f(y)$ , equate $y - 2$ $y-2$ to zero and make $y$ $y$ the subject of the formula

$y - 2 = 0$ $y-2=0$

$y = 2$ $y=2$

Equate $f (y)$ $f(y)$ to zero and substitute for $y = 2$ $y=2$

$y^{2} - P y - 10 = 0$ $y^2-Py-10=0$

$2^{2} - 2 P - 10 = 0$ $2^2-2P-10=0$

$4 - 10 - 2 P = 0$ $4-10-2P=0$

$- 6 - 2 P = 0$ $-6-2P=0$

Make $P$ $P$ the subject of the formula

$- 6 - 2 P = 0$ $-6-2P=0$

$- 2 P = 6$ $-2P=6$

$P = \frac{6}{- 2}$ $P=6/(-2)$

$P = - 3$ $P=-3$