How do you find the zeros by rewriting the function #y=2x^2-578# in intercept form?
1 Answer
Jul 20, 2017
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#
