Question

Find all the zeros of 2x4+x3-14x2-19x-6 two of its zeros are -2 and -1

Answer 1

Answer:

$-2,\ -1,\ -\dfrac{1}{2},\ 3.$

Step-by-step explanation:

Consider polynomial $2x^4+x^3-14x^2-19x-6.$

If x=-2 is its zero, then you can divide the polynomial $2x^4+x^3-14x^2-19x-6$ by $x+2$ and get

$2x^4+x^3-14x^2-19x-6=(x+2)(2x^3-3x^2-8x-3).$

If x=-1, then the polynomial $2x^4+x^3-14x^2-19x-6$ can be rewritten as

$2x^4+x^3-14x^2-19x-6=(x+2)(x+1)(2x^2-5x-3).$

The quadratic polynomial has roots

$x_{1,2}=\dfrac{-(-5)\pm\sqrt{(-5)^2-4\cdot2\cdot(-3)}}{2\cdot 2}=\dfrac{5\pm \sqrt{25+24}}{4}=\dfrac{5\pm \sqrt{49}}{4}=3,-\dfrac{1}{2}.$

Then the polynomial $2x^4+x^3-14x^2-19x-6$ has zeros $-2,\ -1,\ -\dfrac{1}{2},\ 3.$