Answer:
[tex](a) x^2-2x=x \times (x+2) \\\\(b) 6x^2+12x=2\times 3 \times x \times ( x + 2) \\\\(c) 3x^3 - 9x=3\times x \times (x^2 -3) \\\\(d) 4x^2 +28 x^3 = 2 \times 2 \times x \times x \times ( 1+ 7x)[/tex]
Step-by-step explanation:
To factorize, at first, do factorize all the terms then take the common terms out.
(a) [tex]x^2-2x[/tex]
This can be written as
[tex]x^2-2x= x\times x + 2 \times x[/tex]
Now, taking x commom from both the terms.
[tex]x^2-2x=x \times (x+2)[/tex]
(b) [tex]6x^2+12x[/tex]
[tex]=2\times 3 \times x\times x + 2\times 2 \times 3 \times x[/tex]
Now, taking [tex]2\times 3\times x[/tex] common from both the terms, we have
[tex]6x^2+12x=2\times 3 \times x \times ( x + 2)[/tex]
This is the required factorization.
(c) [tex]3x^3 - 9x \\\\[/tex]
[tex]= 3 \times x \times x \times - 9\times x \\\\ 3x^3 - 9x=3\times x \times (x^2 -3)[/tex]
This is the required factorization.
(d) [tex]4x^2 +28 x^3[/tex]
[tex]= 2 \times 2 \times x \times x + 2 \times 2 \times 7 \times x \times x \times x \\\\4x^2 +28 x^3 = 2 \times 2 \times x \times x \times ( 1+ 7x)[/tex]
This is the required factorization.