If xy=0 we assume x and y equal 0
so
the zeros are wehre f(x)=0
0=5(2x-5)(5x+4)
set each to zero
5 is not equal 0 so we don't do that
0=2x-5
5=2x
5/2=x
0=5x+4
-4=5x
-4/5=x
zeroes at x=5/2 and -4/5
Answer: Adiya’s method is not correct. To form a perfect square trinomial, the constant must be isolated on one side of the equation. Also, the coefficient of the term with an exponent of 1 on the variable is used to find the constant in the perfect square trinomial. Adiya should first get the 20x term on the same side of the equation as x2. Then she would divide 20 by 2, square it, and add 100 to both sides.
Answer:
Step-by-step explanation:
Expand:
Add 27 to both sides:
Add 12x to both sides:
Divide both sides by 4:
Answer:
3x³ - 2x² + 1 = 0
Step-by-step explanation:
By definition, a quadratic equation cannot have an exponent higher than 2.
