Answer: [D]: "
x = 0, -3, 5/2 "
.
{Assuming: "
2x³ + x² − 15x = 0 ".
}.
__________________________________________Explanation:___________________________________Given:
_________________________________ 2x³ + x² − <span>15x ;
__________________________________</span> → (2x³ + x²) − 15x ;
→ 2x³ + x² − 15x = x * (2x² + x − 15) ;
→ Factor the expression: "(2x² + x − 15)" ;
→ (2x² + x − 15) =
→ 2x² − 5x + 6x − 15 ;
→ Add the "first TWO (2) terms", and pull out the "like factors" ; → x*(2x − 5) ;
→ Add the "last TWO (2) terms, pulling out "common factors";
→ 3*(2x − 5) ;
→ Now, add up the previous FOUR (4) terms; to get:
→ (x + 3)(2x − 5) ;
→ Now, we have factored the:
"(2x² + x − 15)" of: " x*(2x² + x − 15) " ;
→ So we add the "x"; and write the entire factored expression:
______________________________________________________ → x*(x + 3)(2x − 5) ;
_____________________________________Now, assume the question is asking to solve for "x" by factoring;
when the expression is "equal to zero" ;
_______________________________________That is, when:
→ x*(x + 3)(2x − 5) = 0 ;
____________________________________________Since we have THREE (3) multiplicands; and anything times "0" equals "0" ;
this equation holds true when:
________________________________ 1) x = 0 ;
_______________ 2) (x + 3) = 0 ;
Subtract "3" from each side of the equation;
x + 3 − 3 = 0 − 3 ;
x = -3 ;
_______________________3) 2x − 5 = 0 ;
Add "5" to each side of the equation;
2x − 5 + 5 = 0 + 5 ;
2x = 5 ; Now, divide EACH side of the equation by "2" ; to isolate "x" on one side of the equation; and to solve for "x" ;
2x/2 = 5/2 ;
x = 5/2; or, write as 2.5; or write as 2<span>½ .
</span>
_____________________________________Put simply, the equation is true when:
_________________________________________ x = 0, -3, 5/2 ; which is: Answer choice: [D]._________________________________________
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