Ask Question
5 February, 01:33

How does this polynomial identity work on numerical relationships?

(y + x) (ax + b)

+1
Answers (1)
  1. 5 February, 01:56
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “How does this polynomial identity work on numerical relationships? (y + x) (ax + b) ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers