All categories

3 years ago

What is the greatest common factor of the terms of the polynomial 12x^4 - 6x^3 + 9x^2

Mathematics

1 answer:

choli [55]3 years ago

3 0

Answer:

$3x^{2}$ is the greatest common factor of the polynomial.

Step-by-step explanation:

Given:

The polynomial is $12x^{4} - 6x^{3} + 9x^{2}$

In order to determine the greatest common factor of all the terms, we find the greatest common factor of the coefficients separately and greatest common factor of the variables separately.

So, factors of 12 = 1, 2, 3, 4, 6, 12

Factors of 6 = 1, 2, 3, 6

Factors of 9 = 1, 3, 9

Therefore, the greatest common factor among 12, 6, and 9 is 3.

Now, factors of $x^{4}$ = $1, x, x^{2}, x^{3}, x^{4}$

Factors of $x^{3}$ = $1, x, x^{2}, x^{3}$

Factors of $x^{2}$ = $1, x, x^{2}$

Therefore, the greatest common factor among $x^{4}, x^{3},x^{2}$ is $x^{2}$ .

Hence, the greatest common factor of all the terms of the polynomial is $3x^{2}$

You might be interested in

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago

PLEASE HELP ME IM STRUGGLING!!!

Kitty [74]

Answer:

The required answer is $c=7\sqrt{3}$