Answer:
See explanation
Step-by-step explanation:
Consider the polynomial [tex]y=x^4+3x^3-27x^2+13x+42[/tex] 4th power polynomial function has at most 4 zeros.
Integer zeros can be only among the divisors of 42:
[tex]\pm1, \ \pm 2,\ \pm 3,\ \pm 6,\ \pm 7,\ \pm 14,\ \pm 21,\ \pm 42[/tex]
Check them:
[tex]y(1)=1^4+3\cdot 1^3-27\cdot 1^2+13\cdot 1+42=1+3-27+13+42=32\neq 0\\ \\y(-1)=(-1)^4+3\cdot (-1)^3-27\cdot (-1)^2+13\cdot (-1)+42=1-3-27-13+42=0\\ \\y(2)=2^4+3\cdot 2^3-27\cdot 2^2+13\cdot 2+42=16+24-108+26+42=0\\ \\y(3)=3^4+3\cdot 3^3-27\cdot 3^2+13\cdot 3+42=81+81-243+39+42= 0\\ \\y(-7)=(-7)^4+3\cdot (-7)^3-27\cdot (-7)^2+13\cdot (-7)+42=2,401-1,029-1,323-91+42= 0[/tex]
Thus,
[tex]y=(x+7)(x+1)(x-2)(x-3)[/tex]
Zeros are plotted in attached diagram.
