Answer:
Zeros are x={-5, -1, -5/2}
Step-by-step explanation:
In order to do this problem via factoring there are tricks one of which is that you have to try to break things down in either (x+1) or (x-1). I tried out to get (x+1) first and it worked. You break numbers down, separate them and factor them and so I obtained:
Since we want to find the zero's meaning roots/x-intercepts y = 0 and so:
[tex]2x^3+17x^2+40x+25=0\\[/tex]
For now we'll only focus on the left hand side of the equation and so:
[tex]2x^3+17x^2+40x+25\\\\x^3+x^3+x^2+16x^2+40x+25\\\\(x^3+x^2)+(x^3+x^2)+15x^2+40x+25\\\\(x^3+x^2)+(x^3+x^2)+(x^2+x)+14x^2+39x+25\\[/tex]
For the last part [tex]14x^2+39x+25[/tex] we can factor this out by first multiplying the outer terms and try to see if any of those factors equal the middle term. Thus, knowing that 25 × 14 = 350 and that 25+14 = 39 we obtain:
[tex](x^3+x^2)+(x^3+x^2)+(x^2+x)+14x^2+14x+25x+25\\\\(x^3+x^2)+(x^3+x^2)+(x^2+x)+14x(x+1)+25(x+1)\\\\x^2(x+1)+x^2(x+1)+x(x+1)+14x(x+1)+25(x+1)\\\\(x^2+x^2+x+14x+25)(x+1)\\\\(2x^2+15x+25)(x+1)\\\\(2x^2+10x+5x+25)(x+1)\\\\(2x(x+5)+5(x+5))(x+1)\\\\(2x+5)(x+5)(x+1)=0[/tex]
Note that 25 × 2 = 50 and that 10 + 5 = 15 (middle term of the polynomial).
Now we have 3 cases (2x+5)=0 , (x+5)=0 and (x+1)=0. By solving for x we obtain:
Case 1:
[tex]2x+5=0\\\\2x=-5\\\\x=-\frac{5}{2}\\[/tex]
Case 2:
[tex]x+5=0\\\\x=-5[/tex]
Case 3:
[tex]x+1=0\\\\x=-1[/tex]
Therefore the zeros for this polynomial are:
x= {-5, -1, -5/2}
