Find two consecutive positive integers such that the square of the smaller increased by 4 times the larger is 64

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

step-by-step explanation:

i'm not sure how to show the work for this

answer : c and d or 4(29 + 6)

4(29) + 4(6)

option (a)

step-by-step explanation:

step 1:   take 2 number whose product equal to 25..

and sum of those two number are -10

step: 2 for example here if we take 5,5 are the number,,

5*5=25

and

5+5=10

but we need -10

so if we take   -5,-5

so -5*-5=25

and -5+(-5)=-10

s0

come the equation

replace -10x by these number

like -5x -5x+25

take common

x(x-5)-5(x-5)=0

(x-5)(x-5)=0