Question

Find the value of below algebriac expression for x=2,y=-1 and z=2
i. 4x2−3y2+5z24x2−3y2+5z2
ii. 3x3−2x(4yz+5x2)3x3−2x(4yz+5x2)
iii. 1−4x(yz+3xy)1−4x(yz+3xy)

Answer 1

Step-by-step explanation:

Find the value of below algebriac expression for x=2,y=-1 and z=2

i. 4x2−3y2+5z24x2−3y2+5z2

ii. 3x3−2x(4yz+5x2)3x3−2x(4yz+5x2)

iii. 1−4x(yz+3xy)1−4x(yz+3xy)

Answer 2

Given,

The values of variables,

$x=2\\y=-1\\z=2$

To find,

Values of the given algebraic expressions at the given values of variables.

$4x^{2} }-3y^{2}+5z^{2}\\3x^{3}-2x(4yz+5x^{2})\\1-4x(yz+3xy)$

Solution,

i) For equation,

$4x^{2}-3y^{2}+5z^{2}$

Let us put the respective values in the above equation,

⇒$4(2^{2})-3((-1)^{2})+5(2^{2})$

⇒$4(4)-3(1)+5(4)$

⇒ $16-3+20$

⇒$33$

Answer is 33.

ii) For equation,

$3x^{3}-2x(4yz+5x^{2})$

Let us put the values,

⇒$3(2^{3})-2(2)[4(-1)(2)+5(2^{2})]$

⇒$3(8)-4[-8+5(4)]$

⇒$24-4[-8+20]$

⇒$24-4(12)$

⇒ $24-48$

⇒$-24$

Answer is -24.

iii) For equation,

$1-4x(yz+3xy)$

Let us put the values,

⇒$1-4(2)[(-1)(2)+3(2)(-1)]$
⇒$1-8[-2-6]$

⇒$1-8(-8)$

⇒$1+64$

⇒$65$

Hence, answer is 65.