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3 years ago

A^3+4a^2+4a factor completely

1 answer:

8 0

So firstly, factor out the GCF, or greatest common factor. To find the GCF, list the factors of every term and the greatest one they all share is their GCF. In this case, the GCF is "a":

$a(a^2+4a+4)$

Next, we are going to apply the square of binomials rule in this situation, which is $(x+y)(x+y)=x^2+2xy+y^2$ In this case:

$a^2+4a+4=(a+2)(a+2)\\\\a(a+2)(a+2)$

Your final answer is $a(a+2)(a+2)$ 

You might be interested in

Find the product and write the answer as a trinomial in standard form (x+2y) (3x-5y)

kodGreya [7K]

Answer:

3x^2 + xy - 10y^2

Step-by-step explanation:

1. Use the FOIL method

3x ^2 − 5xy + 6yx − 10y ^2

2. Collect like terms

3x ^2 + (−5xy + 6xy) − 10y^2

3. Simplify

3x ^2 + xy − 10y^2

8 0

4 years ago

Is x+y+1=0 a tangent of both y^2=4x and x^2=4y parabolas?

Lubov Fominskaja [6]

Answer:

yes

Step-by-step explanation:

The line intersects each parabola in one point, so is tangent to both.

For the first parabola, the point of intersection is ...

y^2 = 4(-y-1)

y^2 +4y +4 = 0

(y+2)^2 = 0

y = -2 . . . . . . . . one solution only

x = -(-2)-1 = 1

The point of intersection is (1, -2).

For the second parabola, the equation is the same, but with x and y interchanged:

x^2 = 4(-x-1)

(x +2)^2 = 0

x = -2, y = 1 . . . . . one point of intersection only

___

If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.

_____

Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.

7 0

3 years ago

Who wants my ip its free of charge

Vikentia [17]

Answer:

hope it's help you ok have a good day

5 0

3 years ago

A ball is launched upward from a height 40 feet above ground level. The ball’s height at t seconds is given by -16t^2=128=40 .

Nesterboy [21]

Answer:

The ball will be at 100 feet at time x=0.5 sec and at time x=7.5 sec

Step-by-step explanation:

The correct question is

A ball is launched upward from a height 40 feet above ground level. the ball's height at t seconds is given by h(t)=-16t^2+128t+40 At what time(s) will the ball be at 100 feet?

we have

$h(t)=-16t^{2}+128t+40$