Question

(viii) 4x-2x2-3x2+x
$4 {x}^{2} - 2 {x}^{2} - 3 {x}^{2} + {x}^{2}$
​

Answer 1

Answer:

0

Step-by-step explanation:

4$x^{2}$ - 2$x^{2}$ - 3$x^{2}$ + $x^{2}$

= (4$x^{2}$ -2$x^{2}$) - (3$x^{2}$ + $x^{2}$)

= 2$x^{2}$ - 2$x^{2}$

= 0 $x^{2}$

= 0

Please mark me brailiest!!

Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

$4 {x}^{2} - 2 {x}^{2} - 3 {x}^{2} + {x}^{2}$

$= > 2 {x}^{2} - 3 {x}^{2} + {x}^{2}$

$= > - {x}^{2} + {x}^{2} = 0$

We get 0 .

Alternatively method :

$4 {x}^{2} - 2 {x}^{2} - 3 {x}^{2} + {x}^{2}$

Taking 2x square common &x square:

$2 {x}^{2} (2 - 1) + {x}^{2} ( - 3 + 1)$

$= > 2 {x}^{2} (1) + {x}^{2} ( - 2)$

$= > 2 {x}^{2} - 2 {x}^{2} = 0$

Learn more :

• Above equation is in the form of Binomial polynomial,as it having the highest power 2.