How does this polynomial identity work on numerical relationships?
(y + x) (ax + b)

Let us take 'a' in the place of 'y' so the equation becomes

(y+x) (ax+b)

Step-by-step explanation:

Step 1:

(a + x) (ax + b)

Step 2: Proof

Checking polynomial identity.

(ax+b )(x+a) = FOIL

(ax+b)(x+a)

ax^2+a^2x is the First Term in the FOIL

ax^2 + a^2x + bx + ab

(ax+b)(x+a)+bx+ab is the Second Term in the FOIL

Add both expressions together from First and Second Term  

= ax^2 + a^2x + bx + ab

Step 3: Proof

(ax+b)(x+a) = ax^2 + a^2x + bx + ab

Identity is Found .

Trying with numbers now

(ax+b)(x+a) = ax^2 + a^2x + bx + ab

((2*5)+8)(5+2) =(2*5^2)+(2^2*5)+(8*5)+(2*8)

((10)+8)(7) =(2*25)+(4*5)+(40)+(16)

(18)(7) =(50)+(20)+(56)

126 =126

divide 5 by 2; 2 will go into five 2 times with a remainder of 1. This means the whole number portion of the mixed number is 2, and the remainder of the fraction is 1; this gives us 2 1/2. To plot 2 1/2, plot a point halfway between 2 and 3.

Step-by-step explanation: