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Sagot :
You can factor out a 5 from the equation since all of the numbers are divisible by 5
5(x^2-x-6=0)
Factor into two binomials
5(x-3)(x+2)=0
Because of the zero product property, we know that one of the two binomials (in parenthesis) must equal 0.
x-3=0 and x+2=0
Solve each INDIVIDUALLY to find the values of x
x= 3 AND x= -2
5(x^2-x-6=0)
Factor into two binomials
5(x-3)(x+2)=0
Because of the zero product property, we know that one of the two binomials (in parenthesis) must equal 0.
x-3=0 and x+2=0
Solve each INDIVIDUALLY to find the values of x
x= 3 AND x= -2
[tex] 5x^2-5x-30=0\\
5(x^2-x-6)=0\\
5(x^2-3x+2x-6)=0\\
5(x(x-3)+2(x-3))=0\\
5(x+2)(x-3)=0\\
x=-2 \vee x=3
[/tex]
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