How do you find the zeros by rewriting the function #y=2x^2-578# in intercept form?

1 Answer
Jul 20, 2017

Answer:

#x=+-17#

Explanation:

#"take out a "color(blue)"common factor"" of 2"#

#rArry=2(x^2-289)#

#x^2-289" is a "color(blue)"difference of squares"#

#["note "289=17^2]#

#rArry=2(x-17)(x+17)larrcolor(red) "in-intercept form"#

#"to obtain the zeros set y = 0"#

#rArr2(x-17)(x+17)=0#

#"set each factor to zero and solve for x"#

#x-17=0rArrx=17#

#x+17=0rArrx=-17#

#rArr" the zeros are " x=+-17#