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Destinyyyy

  • one year ago

Solve the polynomial equation by factoring and then using the zero-product principle.

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  1. Destinyyyy
    • one year ago
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    x^3 +x^2 = 4x+4

  2. anonymous
    • one year ago
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    start by making one side 0 by subtracting 4x and 4 from both sides

  3. anonymous
    • one year ago
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    \[x^3+x^2−4x−4=0\] Now group the terms and factor out the gcf from each group \[(x^3+x^2)+(−4x−4)=0\]

  4. Destinyyyy
    • one year ago
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    Okay

  5. anonymous
    • one year ago
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    what's the gcf of x³ and x²?

  6. Destinyyyy
    • one year ago
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    x

  7. anonymous
    • one year ago
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    not just x. They both have x². What do you have inside the parentheses if you factor out x²?

  8. Destinyyyy
    • one year ago
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    Um x

  9. anonymous
    • one year ago
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    Not just x. There are 2 terms originally, so there will still be 2 terms after you factor. let's try it this way |dw:1440773157239:dw|

  10. anonymous
    • one year ago
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    Reduce the part in parentheses to get the factor

  11. Destinyyyy
    • one year ago
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    ? By adding the factions?

  12. Destinyyyy
    • one year ago
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    fractions *

  13. anonymous
    • one year ago
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    no, you're not adding them, just reducing them individually. Like x³/x² = x. What's x²/x² ?

  14. Destinyyyy
    • one year ago
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    1

  15. anonymous
    • one year ago
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    right so x³ + x² in factored form is \[x^2(x + 1)\]

  16. anonymous
    • one year ago
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    Follow the same idea to factor \[(-4x-4)\] First find the gcf of -4x and -4, then divide by it to find what's left in the parentheses

  17. Destinyyyy
    • one year ago
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    -4(x+1)

  18. anonymous
    • one year ago
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    right so the whole equation is now \[x^2(x+1)-4(x+1)=0\] Now we can factor out the common term (x + 1) to get \[(x+1)(x^2-4)=0\] And the x² - 4 part can be factored using the difference of squares rule

  19. Destinyyyy
    • one year ago
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    Um whats that?

  20. Destinyyyy
    • one year ago
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    the difference of squares rule?

  21. anonymous
    • one year ago
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    \[a^2-b^2=(a-b)(a+b)\] you use it to factor when both terms are perfect squares and they are being SUBTRACTED. It doesn't work with addition. x² and 4 are both perfect squares, so x² - 4 factors to (x - 2)(x + 2)

  22. anonymous
    • one year ago
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    So now your equation is \[(x + 1)(x-2)(x+2)=0\] Set each of those factor equal to 0 and solve for x to get your solutions

  23. Destinyyyy
    • one year ago
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    x= -1,2,-2

  24. anonymous
    • one year ago
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    yes

  25. Destinyyyy
    • one year ago
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    Okay thank you

  26. Destinyyyy
    • one year ago
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    Can you help me with another one?

  27. anonymous
    • one year ago
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    i'm actually about to leave now sorry

  28. Destinyyyy
    • one year ago
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    Oh.. Thats okay

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