Question

√3x²+11×+6√3 please do it

Answer 1

√3x^2+11x+6√3

By splitting middle term

√3x^2+9x+2x+6√3

Taking common

√3x(x+3√3)+2(x+3√3)

(√3x+2) (x+3√3)

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Answer 2

Factors of polynomial are

$\bf \red{\sqrt{3} {x}^{2} + 11x + 6 \sqrt{3} = (\sqrt{3}x + 2)( {x} +3 \sqrt{3})}$

Given:

• $\sqrt{3} {x}^{2} + 11x + 6 \sqrt{3}$

To find:

• Factorise the quadratic polynomial.

Solution:

Concept to be used:

Split the middle term so that it's addition/subtraction will be equal to multiplication of outer terms.

Step 1:

Middle term is 11x

Multiplication of outer terms:$\sqrt{3} {x}^{2} \times 6 \sqrt{3}$

or

$= 6 \times 3 {x}^{2}$

or

$= 18 {x}^{2}$

So,

$\bf 11x = 9x + 2x$

as

$9x \times 2x = 18 {x}^{2}$

Step 2:

Factorise the polynomial.

$\sqrt{3} {x}^{2} + 9x + 2x + 6 \sqrt{3}$

or

9=3×3 and 3=√3×√3

so,

$\sqrt{3} {x}^{2} + \sqrt{3} \times \sqrt{3} \times 3 x + 2x + 2 \times 3 \sqrt{3}$

or

$\sqrt{3}x( {x} +3 \sqrt{3}) + 2(x + 3 \sqrt{3} )$

or

$(\sqrt{3}x + 2)( {x} +3 \sqrt{3})$

Thus,

$\bf \sqrt{3} {x}^{2} + 11x + 6 \sqrt{3} = (\sqrt{3}x + 2)( {x} +3 \sqrt{3})$

Learn more:

1) factorise by splitting the middle term :2x^2-7x-15

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2) 2√2x^2+9x+5√2 by splitting the middle term factor

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