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Destinyyyy
 one year ago
Solve the polynomial equation by factoring and then using the zeroproduct principle.
Destinyyyy
 one year ago
Solve the polynomial equation by factoring and then using the zeroproduct principle.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0start by making one side 0 by subtracting 4x and 4 from both sides

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x^3+x^2−4x−4=0\] Now group the terms and factor out the gcf from each group \[(x^3+x^2)+(−4x−4)=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what's the gcf of x³ and x²?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not just x. They both have x². What do you have inside the parentheses if you factor out x²?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not 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
 one year ago
Best ResponseYou've already chosen the best response.0Reduce the part in parentheses to get the factor

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0? By adding the factions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, you're not adding them, just reducing them individually. Like x³/x² = x. What's x²/x² ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right so x³ + x² in factored form is \[x^2(x + 1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Follow the same idea to factor \[(4x4)\] First find the gcf of 4x and 4, then divide by it to find what's left in the parentheses

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right 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 \[(x+1)(x^24)=0\] And the x²  4 part can be factored using the difference of squares rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the difference of squares rule?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a^2b^2=(ab)(a+b)\] 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
 one year ago
Best ResponseYou've already chosen the best response.0So now your equation is \[(x + 1)(x2)(x+2)=0\] Set each of those factor equal to 0 and solve for x to get your solutions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you help me with another one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm actually about to leave now sorry
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