Question:

(-8x³+4x²-6x+9) - (15x³+3x²-8x-17)?

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i know when you add exponents you're supposed to multiply the exponents and when you multiply you add the exponents. what do you do when you subtract exponents? please help me find the answer!

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8 ANSWERS


  1. Timboy had it close.

    -23x^3 + x^2 +2x + 26


  2. you just add or subtract the terms which has the same power or exponents...

    the answer to your question is:

            -23X^3 + X^2 + 2x + 26

    or

            -(23X^3 - X^2 -2x -26)

  3. VERY SIMPLE . EXPAND THE BRACKETS, AND GROUP LIKE TERMS INTERMS OF EXPONENT VALUE, MAKE SURE YOU TAKE THEIR RESPECTIVE SIGNS AND THEN SOLVE....

    -8x³+4x²-6x+9 - 15x³-3x²+8x+17

    -8x³- 15x³+4x²-3x²-6x+8x+9+17

    FINAL ANSWER: -23x³+x²+2X+26

  4. First, you distribute the subtraction sign to everything inside the second pair of parenthases.  You can do this because the subtraction sign is actually a negative 1 -- mutiplying everything inside there.  So you are left with:

    -8x³+4x²-6x+9 -15x³-3x²+8x+17

    Then, it's just a matter of combining like terms:

    -8x³ and -15x³ = -23x³

    +4x² and -3x²= +x²

    -6x and +8x= +2x

    +9 and +17= +26

    Then you put each monomial in order, from greatest power to lowest:

    -23x³+x²+2x+26 -- voila, your answer.

    Hope, I helped.

  5. divide i think... sorry i learned this last year.... but im only in 8th grade! Good luck tho nd im pretty sure its divide:-)

  6. -8x³ + 4x² - 6x + 9 - 15x³ - 3x² + 8x + 17

    -23x³ + 1x² + 2x + 28


  7. -23x^3 + x^2 +2x - 8

  8. All you need to do is collect like terms. Meaning, in this equation, for example -8x^3 and -15x^3.  (Do not forget that the subtraction changes the sign if all the operations inside the parentheses)  Now, you DO NOT do anything with the exponents in addition and subtraction, you simply do the adding or subtracting.  In order to add or subtract, the values must have the same power.  Therefore, you have -8x^3 - 15x^3 = -23x^3.  Continuing do this with all of the terms to get your answer.

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