Ask Question
19 July, 20:24

How does this polynomial identity work on numerical relationships?

(y + x) (ax + b)

+3
Answers (1)
  1. 19 July, 22:22
    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 you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers