Factor the greatest common factor out from polynomial.. 14x2 + 49x

24x^3 - 16x^2 + 68x

4x would be ur GCF

[tex]4x^2+y^2+4xy+8x+4y+4\\\\=\underbrace{(2x)^2+2\cdot2x\cdot y+y^2}_{(a+b)^2=a^2+2ab+b^2}+8x+4y+4\\\\=\underbrace{(2x+y)^2+2(2x+y)\cdot2+2^2}_{(a+b)^2=a^2+2ab+b^2}\\\\=[(2x+y)+2]^2=(2x+y+2)^2[/tex]

  No. We can conclude that (x-4) is not a factor of the polynomial.

Step-by-step explanation:

The polynomial remainder theorem tells you (x-4) will be a factor of the polynomial f(x) if and only if f(4) = 0. When we evaluate ...

  f(x) = 4x³ -20x² +18x -12

at x=4, we find that f(4) = -4. This is not zero, so (x-4) is not a factor of f(x).

_____

The evaluation is perhaps easiest done by hand when the function is written in Horner form:

  f(x) = ((4x -20)x +18)x -12

  f(4) = ((4·4 -10)4 +18)4 -12 = ((-4)4 +18)4 -12 = 2·4 -12

  f(4) = -4

_____

The one real factor of this polynomial is irrational, and the remaining factors are complex and irrational.

3x

2x-1

4x^2 + 2x +1

Step-by-step explanation:

[tex]24x^4- 3x[/tex]

We take out GCF 3x. divide each term by 3x

[tex]3x(8x^3 -1)[/tex]

8x^3 -1 can be written as (2x)^3 - 1^3

Now we apply a^3 - b^3 formula

[tex]a^3 - b^3 = (a-b) (a^2+ ab+b^2)[/tex]

[tex](2x)^3 - 1^3 = (2x-1) ((2x)^2+ (2x)(1)+1^2)[/tex]

[tex]=(2x-1)(4x^2+2x+1)[/tex]

[tex]24x^4- 3x=3x(2x-1)(4x^2+2x+1)[/tex]

So prime factors are

3x

2x-1

4x^2 + 2x +1

Step-by-step explanation:

does it have to be right?

The answer is true(apex verified)

Step-by-step explanation:

The right answers for this problem would be

B. 3x

D. (2x-1)

F. (4x^2 + 2x + 1)

Hope this helps! :3

YES!!!

Step-by-step explanation:

When I factor [tex]4x^2-6x-4[/tex],

I get: [tex]2\left(x-2\right)\left(2x+1\right)[/tex]

Notice the [tex]\left(x-2\right)[/tex],

The polynomial [tex](x-2)[/tex] is a factor of the polynomial  [tex]4x^2-6x-4[/tex]

Hope this helps!!

True, x-2 is a factor

You can take the square root of everything there and it leads to this.

(2x-4)(2x+4)