Question

In an art & craft class Sameer learnt how to paint walls of a room. He practiced a lot and tried it on small models. Finally, he decided to paint his room at home. The room was cuboidal in shape, with length being the longest and height the shortest. If the volume of the room can be written in a polynomial is x3+6x2+3x-10, then: i.) The length of cuboid in polynomial is ii.) If x=2, then the area of floor is iii.) If x=8, then its height is iv.) What type of polynomial the volume is?


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Answer 1

SOLUTION

GIVEN

In an art & craft class Sameer learnt how to paint walls of a room. He practiced a lot and tried it on small models. Finally, he decided to paint his room at home. The room was cuboidal in shape, with length being the longest and height the shortest.

The volume of the room can be written in a polynomial is x³ + 6x² + 3x - 10

TO DETERMINE

i) The length of cuboid in polynomial

ii) If x = 2 then the area of floor

iii) If x = 8, then its height is

iv) What type of polynomial the volume

EVALUATION

Here it is given that the volume of the room can be written in a polynomial is x³ + 6x² + 3x - 10

Now

$\sf{ {x}^{3} + 6 {x}^{2} + 3x - 10 }$

$\sf{ = {x}^{3} - {x}^{2} + 7 {x}^{2} - 7x+ 10x - 10 }$

$\sf{ = {x}^{2}(x - 1) + 7x(x - 1) + 10(x - 1)}$

$\sf{ = (x - 1)({x}^{2} + 7x + 10)}$

$\sf{ = (x - 1)({x}^{2} + 5x + 2x+ 10)}$

$\sf{ = (x - 1)(x + 2)(x + 5)}$

Thus we have

Length = x + 5

Breadth = x + 2

Height = x - 1

(i) Length of the cuboid in polynomial

= ( x + 5 ) unit

(ii) Area of the floor

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

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

$\sf{ = {x}^{2} +7x + 10 }$

For x = 2 the area of the floor

$\sf{ = {2}^{2} +(7 \times 2) + 10 }$

$\sf{ = 4 + 14 + 10 }$

$\sf{ =28}$

(iii) If x = 8 we have

Height

= 8 - 1 unit

= 7 unit

(iv) In the polynomial the highest power of its variable that appears with nonzero coefficient is 3

Hence the polynomial is a cubic polynomial

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