Question

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Factor completely 12x4 + 6x3 + 18x2.

Prime
3(4x4 + 2x3 + 6x2)
3x2(4x2 + 2x + 6)
3x(4x3 + 2x2 + 6x)


Factor completely 7x3y +14x2y3 − 7x2y2.

7x2(xy + 2y3 – y2)
7x2y(x + 2y2 – y)
7(x3y +2x2y3 − x2y2)
Prime


What is the Greatest Common Factor of x6 and x3?

x4
x3
x2
x

Answer 1

Question 1.) Factor completely $12x^{4}+ 6x^{3}+18x^{2}$

Since $6$ divides all the three terms of the equation, so we take $6$ common out from the equation.

i.e., $6(2x^{4}+ 1x^{3}+3x^{2})$

Now, since $x^{2}$ divides all the three terms of the equation, so we take $x^{2}$ common out from the equation.

i.e., $6x^{2}(2x^{2}+ 1x+3)$

Now, this cannot be factored further as $2x^{2}+ 1x+3$ is a quadratic equation and taking out discriminant of it is:

$D = (1)^2 - 4(2)(3)\\  \ \   = 1- 24\\ \ \ = -23<0$

Since, the discriminant is less than zero. Therefore, the factored form of the given equation :  $12x^{4}+ 6x^{3}+18x^{2}$  is : $6x^{2}(2x^{2}+ 1x+3)$

Question 2.) $7x^{3}y+ 14x^{2} y^{3} - 7x^{2}y^{2}$

Since $7$ divides all the three terms of the equation, so we take $7$ common out from the equation.

i.e., $7(x^{3}y+ 2x^{2} y^{3} - x^{2}y^{2})$

Now, since $x^{2}y$ divides all the three terms of the equation, so we take $x^{2}y$ common out from the equation.

i.e., $7x^{2}y(x+ 2y^{2} - y)$

Therefore, the factored form of the given equation :  $7x^{3}y+ 14x^{2} y^{3} - 7x^{2}y^{2}$  is : $7x^{2}y(x+ 2y^{2} - y)$

Question 3.) What is the Greatest Common Factor of $x^{6}$ and $x^{3}$ ?

The Greatest Common Factor of two numbers is the biggest number which divides both the given numbers completely.

Now, for the expressions $x^{6}$ and $x^{3}$:

Since, $x^{3}$ divides both the expressions completely and it is the biggest expression which divides both the expressions.

Therefore, $x^{3}$ is the Greatest Common Factor of $x^{6}$ and $x^{3}$

Answer 2

Answer:

#2 is definatly B

Step-by-step explanation: