How do you solve #x^4+2x^3=15x^2#?

1 Answer
Mar 5, 2018

Answer:

#x=-5" or "x=0" or "x=3#

Explanation:

#"rearrange equation, equating to zero"#

#x^4+2x^3-15x^2=0#

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

#rArrx^2(x^2+2x-15)=0#

#"factorise the quadratic using the a-c method"#

#"the factors of - 15 which sum to + 2 are + 5 and - 3"#

#rArrx^2(x+5)(x-3)=0#

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

#x^2=0rArrx=0#

#x+5=0rArrx=-5#

#x-3=0rArrx=3#