Which divisor of -2x^3 - 5x^2 + 4x - 2 results in a remainder of -14

Mathematics

1 answer:

Vsevolod [243]3 years ago

3 0

Answer:

$f(x) = -2x^3 -5x^2 +4x -2\\option 1 : x+2 \\f(-2) = -2(-2)^3 -5(-2)^2 +4(-2)-2 = 16 -20-8-2= -14\\\\\\option 2: x+3\\f(-3) = -2(-3)^3 -5(-3)^2 +4(-3) - 2 = 54 -45 -12-2 = -5\\\\option 3: x-2\\f(2) = -2(2)^3-5(2)^2-4(2)-2=-16 -20-8-2=-46\\\\option4 :x-3\\f(3) = -2(3)^3 - 5(3)^2-4(3)-2 = -54-45-12-2=-113\\$

option 1: x + 2

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2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

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<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

<u>Slope:</u>

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

<u>Slope:</u>

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$