solve this ................... Created by caketea. algebra-ru

1.Proof:Let be a given polynomial function.Then, by Polynomial remainder theorem, is a divisor (factor) of if and only if .Hence,Q.E.D.2. Since is a divisor of , there is exists such that .Hence, .Therefore,Now, let's find the roots of g(x):Hence, are solutions of f(x).

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caketea @caketea

July 2022 1 18 Report

solve this ..................

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1.
Proof:
Let $f(x)$ be a given polynomial function.
Then, by Polynomial remainder theorem, $(x-2)$ is a divisor (factor) of $f(x)$ if and only if $f(2)=0$ .

Hence,
$f(2)=2\cdot 8 -9\cdot 4+7\cdot 2+6=16-36+14+6=0$

Q.E.D.

2.
Since $(x-2)$ is a divisor of $f(x)=2x^3-9x^2+7x+6$ , there is exists $g(x)$ such that $f(x)=(x-2)\cdot g(x)$ .
Hence,
$g(x)=f(x)/(x-2)$ .
Therefore,
$g(x)=2x^2-5x-3$

Now, let's find the roots of g(x):
$\displaystyle x_{1,2}= \frac{5\pm \sqrt{25+24} }{4}= \frac{5\pm7}{4}= 3,- \frac{1}{2}$

Hence,

$x_1,x_2,x_3=3,- 0.5,2$ are solutions of f(x).

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