Question

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Answer 1

Answer:

$(a) x^2-2x=x \times (x+2) \\\\(b) 6x^2+12x=2\times 3 \times x \times ( x + 2) \\\\(c) 3x^3 - 9x=3\times x \times (x^2 -3) \\\\(d) 4x^2 +28 x^3 = 2 \times 2 \times x \times x \times ( 1+ 7x)$

Step-by-step explanation:

To factorize, at first, do factorize all the terms then take the common terms out.

(a) $x^2-2x$

This can be written as

$x^2-2x= x\times x + 2 \times x$

Now, taking x commom from both the terms.

$x^2-2x=x \times (x+2)$

(b) $6x^2+12x$

$=2\times 3 \times x\times x + 2\times 2 \times 3 \times x$

Now, taking $2\times 3\times x$ common from both the terms, we have

$6x^2+12x=2\times 3 \times x \times ( x + 2)$

This is the required factorization.

(c) $3x^3 - 9x$

$= 3 \times x \times x \times - 9\times x \\\\ 3x^3 - 9x=3\times x \times (x^2 -3)$

This is the required factorization.

(d) $4x^2 +28 x^3$

$= 2 \times 2 \times x \times x + 2 \times 2 \times 7 \times x \times x \times x \\\\4x^2 +28 x^3 = 2 \times 2 \times x \times x \times ( 1+ 7x)$

This is the required factorization.