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3 years ago

One zero of f(x)=x^3-2x^2-5x+6 is 1 what are other zeros of the function

Mathematics

2 answers:

Alexus [3.1K]3 years ago

8 0

Answer:

-2 and 3

Step-by-step explanation:

Given one zero of f(x)=x^3-2x^2-5x+6 to be 1 this means that x-1 is a factor of the polynomial x³-2x²-5x+6

Before we can get the other two factors. We need to divide the polynomial function by x-1 and factorize the resulting quotient as shown in the attachment.

Quotient Q(x) = x²-x-6

Factorizing this function

x²-3x+2x-6 = 0

x(x-3)+2(x-3) = 0

(x+2)(x-3) = 0

x+2 = 0 and x-3 = 0

x = -2 and x= 3

Therefore the other zeros of the function are -2 and 3

elena-s [515]3 years ago

7 0

Answer:

The other zeros are -2 and 3.

Step-by-step explanation:

As 1 is a zero then (x - 1) is a factor so we divide:

x - 1 )x^3-2x^2-5x+6(x^2-x-6

x^3-x^2

-x^2-5x

-x^2+x

-6x+6

...........

x^2 - x - 6 = 0

(x - 3)(x + 2)

x = -2, 3.

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Alex777 [14]

Answer:

3/3 or just 3.

Step-by-step explanation:

everytime the line crosses the x and y coordinates is basically a plot. so counting from the bottom left hand corner to the "plot" in the middle is 3/3 because it rises three and runs positive 3.

8 0

3 years ago

Find the surface area of a sphere with a volume of 36π in3.

swat32

First use known volume to solve for radius:
$V=\frac{4}{3}\pi r^{3}$
$r=\sqrt[3]{\frac{3V}{4\pi}}$
$r=\sqrt[3]{\frac{3(36\pi)}{4\pi}}$
$r=\sqrt[3]{27}$
$r=3$

Then use the radius to get the surface area:
$A_{surf}=4 \pi r^{2}$
$A_{surf}=4 \pi 3^{2}$
$A_{surf}=4 \pi 9$
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6 0

2 years ago

If a certain cannon is fired from a height of 8.8 meters above the ground, at a certain angle, the height of the cannonball ab

Dennis_Churaev [7]

Answer:

It would take approximately 6.50 second for the cannonball to strike the ground.

Step-by-step explanation:

Consider the provided function.

$h(t)=-4.9t^2+30.5t+8.8$

We need to find the time takes for the cannonball to strike the ground.

Substitute h(t) = 0 in above function.

$-4.9t^2+30.5t+8.8=0$

Multiply both sides by 10.

$-49t^2+305t+88=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are: $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute a = -49, b = 305 and c=88

$t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}$