Question

2√3x^2+x-5√3 solve with midle term splitting​

Answer 1

Step-by-step explanation:

Step by step solution :

STEP

1

:

Equation at the end of step 1

(3x2 - 2x) - 5 = 0

STEP

2

:

Trying to factor by splitting the middle term

2.1 Factoring 3x2-2x-5

The first term is, 3x2 its coefficient is 3 .

The middle term is, -2x its coefficient is -2 .

The last term, "the constant", is -5

Step-1 : Multiply the coefficient of the first term by the constant 3 • -5 = -15

Step-2 : Find two factors of -15 whose sum equals the coefficient of the middle term, which is -2 .

-15 + 1 = -14

-5 + 3 = -2 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -5 and 3

3x2 - 5x + 3x - 5

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (3x-5)

Add up the last 2 terms, pulling out common factors :

1 • (3x-5)

Step-5 : Add up the four terms of step 4 :

(x+1) • (3x-5)

Which is the desired factorization

Equation at the end of step

2

:

(3x - 5) • (x + 1) = 0

STEP

3

:

Theory - Roots of a product

3.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Answer 2

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2√3x^2+x-5√3 solve with midle term splitting

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$2 \sqrt{3} {x}^{2} + x - 5 \sqrt{3}$

=>Now , multiplying the coefficient of x^2 with constant:

$= > 2 \sqrt{3} \times 5 \sqrt{3} = 3 \times 5 \times 2 = 30$

=>we obtain 30 we have to think a number whose sum and substract is 1 and it's product becomes 30.

6×5 =30 and 6x-5x=x

$= > 2 \sqrt{3} {x}^{2} + 6x - 5x - 5 \sqrt{3}$

$= > 2 \sqrt{3} x(x + \sqrt{3} ) - 5(x + \sqrt{3} )$

$\bold{= > (2 \sqrt{3x} - 5)(x + \sqrt{3} )}$

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