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ADDING/SUBTRACTING BINOMIALS WITH POLYNOMIALS

ADDING/SUBTRACTING BINOMIALS WITH POLYNOMIALS. Extended Algebra Topics. PREVIOUS KNOWLEDGE. We have been combining monomials with monomials

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ADDING/SUBTRACTING BINOMIALS WITH POLYNOMIALS

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  1. ADDING/SUBTRACTING BINOMIALS WITH POLYNOMIALS Extended Algebra Topics

  2. PREVIOUS KNOWLEDGE • We have been combining monomials with monomials • Even when we did 3x + 7 + 21x – 11, we were actually only combining monomials because 3x and 21x are both monomials (we put them together to get 24x) and +7 and – 11 are both monomials (we get – 4 when we put them together). • So, even though our answer (24x – 4) is a binomial, we really only combined monomials. • Today, we will look at binomials being added or subtracted to other polynomials.

  3. MONOMIAL WITH MONOMIAL • 3x + 9x • 7y + 11y • 9n – 3n • 123p – 65p • We always did either the first term plus the second or the first term minus the second…easy right?

  4. MONOMIAL PLUS BINOMIAL • 3x + (2x + 7) • Continue to do the first part plus the second part…in other words, 3x plus all of 2x + 7 • Combine what can be combined…the 3x and the 2x. Thus, do 3x + 2x to get 5x. • Since the 7 has no one to “team up” with, you just tack him on the end. • Answer: 5x + 7

  5. TRY SOME • 12y + (3y + 90) • 89m + (-3m + 11) • -78y + (-9y – 123) • 99r + (7p – 23) • 45h + (15h – 60)

  6. MONOMIAL MINUS BINOMIAL • BE CAREFUL WITH SUBTRACTION • 15m – (13m + 7) • The parenthesis indicate that you are subtracting the whole 13m + 7 from the 15m. • Thus, you have to do 15m – 13m to get 2m • Since there is no term like 7 in the first part of the problem, you must treat it as 0 – 7, which gets you -7 instead of +7 • It looks like this….15m + 0 – (13m + 7) • So, 15m – 13m is 2m…+ 0 – 7 is -7 • Answer is 2m - 7

  7. TRY SOME • 11m – (3m + 8) • 32x – (12x + 9) • 45p – (11p – 10) • -32k – (5 + 23k) • 88m – (23m – 56)

  8. BINOMIAL PLUS BINOMIAL • (5m + 9) + (3m + 10) • With this situation, the first part has a binomial as does the second part. Thus, you are just combining things that are like • 5m + 3m and +9 + 10 • 8m +19 • Answer is 8m + 19 • Watch your signs…remember the role of the minus sign can change into a negative sign.

  9. TRY SOME • (5m + 6) + (3m + 65) • (11x – 5) + (9x + 13) • 12x + 19 + (-20 – 19x) • (-43p – 56) + (124p + 14) • (-67n – 88) + (-12n – 64)

  10. BINOMIAL MINUS BINOMIAL • (5m + 9) – (2m + 6) • Remember, you have all of the first part, minus all of the second part. • Thus, start with the 5m and subtract the part of the second binomial that can be subtracted from 5m (aka the 2m)….5m – 2m • Then, do the positive 9 – the positive 6 or…9 – 6 • Answer is 3m + 3 • WATCH OUT FOR THE SIGNS!

  11. TRY SOME • (6p + 7) – (2p + 4) • (13x + 9) – (12x + 6) • (15m – 13) – (11m + 4) • (-34r – 20) – (-21r + 30) • (45p + 43) – (45p – 36)

  12. WATCH OUT FOR “SUBTRACTED FROM” • If an example says in words….”What is 5x subtracted from 13x?”, how would you do that? • 13x – 5x • Remember, subtracted from was “a star”, which meant that the numbers need to be flipped around…just like less than. • So, 11x + 9 subtracted from 15x + 12 should be written how? • (15x + 12) – (11x + 9)

  13. TRY SOME • What is 5x + 13 subtracted from 8x + 11? • What is -4x – 9 subtracted from 11x + 5? • What is 15x – 21 subtracted from 12x – 8?

  14. SOMETHING TO HELP WITH THE SUBTRACTION • Remember what we did when we subtracted integers? We did “add the opposite”. In other words, we said that if you change the subtraction sign to addition, and then change the sign on the second integer to it’s opposite, we can proceed with addition. • Let’s apply this to this topic • Example: 32x – (12x + 7) • Changes to 32x + (-12x – 7) • In this case, you change the minus to plus, and you change EVERYTHING in the second part of the problem to it’s opposite. See if you get the same answer as the previous way.

  15. TRY SOME • (5x + 11) – (2x – 9) • (19y – 23) – (9y + 17) • 82m – 9 subtracted from 42m + 99 • -34w – 19 subtracted from -17w – 89

  16. KEEP YOUR EYE OUT FOR FRACTIONS • Remember that you need a common denominator when you add or subtract fractions, but in this case, you will only need it for the common terms. All other rules still apply (subtracted from etc)

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