Question

Plot the zeros of the polynomial y = x4 + 3x3 − 27x2 + 13x + 42

Answer 1

Answer:

See explanation

Step-by-step explanation:

Consider the polynomial $y=x^4+3x^3-27x^2+13x+42$ 4th power polynomial function has at most 4 zeros.

Integer zeros can be only among the divisors of 42:

$\pm1, \ \pm 2,\ \pm 3,\ \pm 6,\ \pm 7,\ \pm 14,\ \pm 21,\ \pm 42$

Check them:

$y(1)=1^4+3\cdot 1^3-27\cdot 1^2+13\cdot 1+42=1+3-27+13+42=32\neq 0\\ \\y(-1)=(-1)^4+3\cdot (-1)^3-27\cdot (-1)^2+13\cdot (-1)+42=1-3-27-13+42=0\\ \\y(2)=2^4+3\cdot 2^3-27\cdot 2^2+13\cdot 2+42=16+24-108+26+42=0\\ \\y(3)=3^4+3\cdot 3^3-27\cdot 3^2+13\cdot 3+42=81+81-243+39+42= 0\\ \\y(-7)=(-7)^4+3\cdot (-7)^3-27\cdot (-7)^2+13\cdot (-7)+42=2,401-1,029-1,323-91+42= 0$

Thus,

$y=(x+7)(x+1)(x-2)(x-3)$

Zeros are plotted in attached diagram.

Answer 2

Answer:

The remainder is 0, so -1 is a zero of the polynomial and x + 1 is another factor: (x − 2)(x + 1)(x2 + 4x − 21). Now factor the remaining polynomial using normal means. The fully factored form is (x − 2)(x + 1)(x + 7)(x − 3), so the zeros of the polynomial are x = 2, -1, -7 and 3.

Step-by-step explanation: