Question 1156960
<br>
The response from the other tutor makes solving the problem harder by multiplying polynomials when not necessary.<br>
{{{x^2-3x = (x^2-3x)^2-3(x^2-3x)}}}<br>
Treat the {{{x^2-3x}}} as a unit (a new "variable"):<br>
{{{0 = (x^2-3x)^2-4(x^2-3x)}}}<br>
Factor out one occurrence of the "variable" from each term:<br>
{{{0 = (x^2-3x)((x^2-3x)-4)}}}
{{{0 = (x^2-3x)(x^2-3x-4)}}}<br>
Now solve by factoring each quadratic.<br>
{{{0 = (x)(x-3)(x-4)(x+1)}}}<br>
ANSWERS: x=0; x=3; x=4; x=-1<br>