One number is 6 times a first number. A third number is 100 more than the first number. If the sum of the three numbers is 812, find the numbers.

Step-by-step explanation:

Let the first number be x

Given second number is 6 times a first number.

If first number is x then second number is 6*x = 6x

Given third number is 100 more than  the first number

If first number is x then third number is x+100 = x+100

Sum of these three numbers in terms of x

is

sum = x+ 6x+x+100 = 8x+100 1

Given, that sum of these numbers is 812.

equating 812 with equation 1 we have,

8x+100  = 812

=> 8x = 812- 100

=> 8x = 712

=> x = 712/8 =  89.

Therefore,

first number = x = 89

second number = 6x = 89*6= 534

third number = x+100 = 89+100 = 189

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form