Question

factorise by using factor theoram 21x^2-x-2​

Answer 1

21x²-7x+6x-2=0

7x(3x-1)+2(3x-1)=0

(7x+2)(3x-1)=0

x = 1/3, -2/7

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Answer 2

$\huge\rm\underline\purple{Answer:-}$

The first term is, 21x2 its coefficient is 21 .

The middle term is, -x its coefficient is -1 .

The last term, "the constant", is -2

Step-1 : Multiply the coefficient of the first term by the constant 21 • -2 = -42

Step-2 : Find two factors of -42 whose sum equals the coefficient of the middle term, which is -1 .

-42 + 1 = -41

-21 + 2 = -19

-14 + 3 = -11

-7 + 6 = -1 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -7 and 6

21x2 - 7x + 6x - 2

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

7x • (3x-1)

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

2 • (3x-1)

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

(7x+2) • (3x-1)

Equation at the end of step

2

:

(3x - 1) • (7x + 2) = 0

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.

• 3x-1 = 0

Add 1 to both sides of the equation :

3x = 1

Divide both sides of the equation by 3:

x = 1/3 = 0.333

• Solve : 7x+2 = 0

Subtract 2 from both sides of the equation :

7x = -2

Divide both sides of the equation by 7:

x = -2/7 = -0.286

hope it's helpful...

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