Destinyyyy
  • Destinyyyy
Solve the polynomial equation by factoring and then using the zero-product principle.
Mathematics
jamiebookeater
  • jamiebookeater
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Destinyyyy
  • Destinyyyy
x^3 +x^2 = 4x+4
anonymous
  • anonymous
start by making one side 0 by subtracting 4x and 4 from both sides
anonymous
  • anonymous
\[x^3+x^2−4x−4=0\] Now group the terms and factor out the gcf from each group \[(x^3+x^2)+(−4x−4)=0\]

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Destinyyyy
  • Destinyyyy
Okay
anonymous
  • anonymous
what's the gcf of x³ and x²?
Destinyyyy
  • Destinyyyy
x
anonymous
  • anonymous
not just x. They both have x². What do you have inside the parentheses if you factor out x²?
Destinyyyy
  • Destinyyyy
Um x
anonymous
  • anonymous
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|
anonymous
  • anonymous
Reduce the part in parentheses to get the factor
Destinyyyy
  • Destinyyyy
? By adding the factions?
Destinyyyy
  • Destinyyyy
fractions *
anonymous
  • anonymous
no, you're not adding them, just reducing them individually. Like x³/x² = x. What's x²/x² ?
Destinyyyy
  • Destinyyyy
1
anonymous
  • anonymous
right so x³ + x² in factored form is \[x^2(x + 1)\]
anonymous
  • anonymous
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
Destinyyyy
  • Destinyyyy
-4(x+1)
anonymous
  • anonymous
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
Destinyyyy
  • Destinyyyy
Um whats that?
Destinyyyy
  • Destinyyyy
the difference of squares rule?
anonymous
  • anonymous
\[a^2-b^2=(a-b)(a+b)\] 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)
anonymous
  • anonymous
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
Destinyyyy
  • Destinyyyy
x= -1,2,-2
anonymous
  • anonymous
yes
Destinyyyy
  • Destinyyyy
Okay thank you
Destinyyyy
  • Destinyyyy
Can you help me with another one?
anonymous
  • anonymous
i'm actually about to leave now sorry
Destinyyyy
  • Destinyyyy
Oh.. Thats okay

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