Question

Factorise 3x^2-3√2x+1 by middle term splitting.

Answer 1

Answer:

To factorize 3x^2-3√2x+1 by center term part, we got to discover two numbers that increase to grant the coefficient of x^2 term (which is 3), and include up to allow the coefficient of x term (which is -3√2).

Step-by-step explanation:

Let us attempt to factorize it step by step:

Step 1:

Duplicate the coefficient of x^2 by the steady term (which is 1), i.e. 3x^2 * 1 = 3x^2.

Step 2:

Discover two numbers whose item is 3x^2 and whose whole is -3√2x.

We will compose 3x^2 as 3x * x.

Ready to type in 1 as the item of two numbers as well, i.e. 1 * 1.

So, the two numbers we require are such that their entirety is -3√2x and their item is 3x.

Let's discover those two numbers:

We are able type in -3√2x as -3√2 * x.

So, we ought to discover two numbers whose item is 3x, and whose whole is -3√2 * x.

The two numbers are -√2 and -3x.

(-√2 * -3x = 3x√2, and -√2 + (-3x) = -3√2x)

Step 3:

Revamp the center term (-3√2x) as the entirety of -√2x and -3x:

3x^2 -3√2x + 1 = 3x^2 - √2x - 3x + 1

Step 4:

Gather the terms as takes after:

(3x^2 - √2x) + (-3x + 1)

Step 5:

Calculate out the common terms from each bunch:

3x(x - 1/√2) - 1(x - 1/√2)

Step 6:

Combine the remaining terms:

(3x - 1)(x - 1/√2)

Subsequently, the given expression 3x^2-3√2x+1 can be factorized as (3x - 1)(x - 1/√2) utilizing center term part.

To learn more about factorize, click on the given link.

https://brainly.in/question/5051642

To learn more about coefficient, click on the given link.

https://brainly.in/question/54991892

#SPJ1

Answer 2

Answer:

$\boxed{\bf \: \dfrac{1}{12}\left[(6x - 3\sqrt{2} + \sqrt{6})x (6x - 3 \sqrt{2} - \sqrt{6})\right] \: }$

Step-by-step explanation:

Given expression is

$\sf \: {3x}^{2} - 3 \sqrt{2}x + 1$

On multiply and divide by 12, we get

$\sf \: = \: \dfrac{1}{12}\left[ {36x}^{2} - 3 6\sqrt{2}x + 12\right]$

$\sf \: = \: \dfrac{1}{12}\left[ {36x}^{2} + 6( - 6\sqrt{2})x + 12\right]$

$\sf \: = \: \dfrac{1}{12}\left[ {36x}^{2} + 6( - 3\sqrt{2} - 3 \sqrt{2} )x + 12\right]$

$\sf \: = \: \dfrac{1}{12}\left[ {36x}^{2} + 6( - 3\sqrt{2} - 3 \sqrt{2} + \sqrt{6} - \sqrt{6} )x + 12\right]$

$\sf \: = \: \dfrac{1}{12}\left[ {36x}^{2} + 6( - 3\sqrt{2} + \sqrt{6} - 3 \sqrt{2} - \sqrt{6})x + 12\right]$

$\sf \: = \: \dfrac{1}{12}\left[ {36x}^{2} + 6( - 3\sqrt{2} + \sqrt{6})x - 6(3 \sqrt{2} + \sqrt{6})x + 12\right]$

$\sf \: = \: \dfrac{1}{12}\left[6x(6x - 3\sqrt{2} + \sqrt{6})x - 6(3 \sqrt{2} + \sqrt{6})x +(3 \sqrt{2} + \sqrt{6})(3 \sqrt{2} - \sqrt{6})\right]$

$\left[ \because \: 12 = 18 - 6 = {(3 \sqrt{2}) }^{2} - {( \sqrt{6}) }^{2} = (3 \sqrt{2} + \sqrt{6})(3 \sqrt{2} - \sqrt{6})\right]$

$\sf \: = \: \dfrac{1}{12}\left[6x(6x - 3\sqrt{2} + \sqrt{6})x - (3 \sqrt{2} + \sqrt{6})(6x - 3 \sqrt{2} + \sqrt{6})\right]$

$\sf \: = \: \dfrac{1}{12}\left[(6x - 3\sqrt{2} + \sqrt{6})x (6x - 3 \sqrt{2} - \sqrt{6})\right]$

Hence,

$\implies\boxed{\sf \: {3x}^{2} - 3 \sqrt{2}x + 1 = \: \dfrac{1}{12}\left[(6x - 3\sqrt{2} + \sqrt{6})x (6x - 3 \sqrt{2} - \sqrt{6})\right] \: }$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$