Question

factorise this.. plzzz give answer ​

Answer 1

Solutions

1. x² - 21x + 90

Splitting the middle term we have

$\sf {x}^{2} - 15x - 6x + 90 \\ = \sf x(x - 15) - 6(x - 15) \\ = \sf (x - 15)(x - 6)$

Thus the factorised form is

(x - 15)(x - 6)

and zeroes will be 15 and 6

______________________

2. x² + 19x - 150

Splitting the middle term we have :

$\sf {x}^{2} - 6x + 25x - 150 \\ = \sf x(x - 6) + 25(x - 6) \\ \sf =( x - 6)(x + 25)$

Thus the factorised form is

(x - 6)(x + 25)

and the zeroes are 6 and -25

______________________

3. x² - 20x - 300

Splitting the middle term we have :

$\sf {x}^{2} + 10x - 30x - 300 \\ \sf = x(x + 10) - 30(x + 10) \\ \sf = (x + 10)(x - 30)$

Thus the factorised form is

(x + 10)(x - 30)

and the zeroes are -10 and 30

______________________

4. x² - 3x - 28

Splitting the middle term we have :

$\sf {x}^{2} - 7x + 4x - 28 \\ \sf = x(x - 7) + 4(x - 7) \\ \sf = (x - 7)(x + 4)$

Thus the factorised form is

(x - 7)(x + 4)

and the zeroes are 7 and -4

_______________________

5. x² - 7x + 10

Splitting the middle term we have :

$\sf {x}^{2} - 2x - 5x + 10 \\ \sf = x(x - 2) - 5(x - 2) \\ \sf = (x - 2)(x - 5)$

Thus the factorised form is

(x - 2)(x - 5)

and the zeroes are 2 and 5

Answer 2

___________________________________

$\huge\tt{GIVEN:}$

1. x²-21x+90
2. x²+19x-150
3. x²-20x-300
4. x²-20x-28
5. x²-7x+10

___________________________________

$\huge\tt{TO~FIND:}$

• Factorising all the equations

___________________________________

$\huge\tt{SOLUTIONS:}$

QUESTION.1

➠x²-15x - 6x + 90

➠x(x-15)-6(x-15)

➠(x-15)(x-16)

___________________________________

QUESTION.2

➠x²+19x-150

➠x²-6x+25x-160

➠x(x-6)+25(x-6)

➠(x-6)(x+25)

___________________________________

QUESTION.3

➠x²-20x-300

➠x²+10x-30x-300

➠x(x+10)-30(x+10)

➠(x+10)(x-30)

___________________________________

QUESTION.4

➠x²-7x+4x-28

➠x(x-7)+4(x-7)

➠(x-7)(x+4)

___________________________________

QUESTION.5

➠x²-2x-5x+10

➠x(x-2)-5(x-2)

➠(x-2)(x-5)

___________________________________