Question

find out zero of p(x) =x2 + 51x - 150
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Answer 1

Answer:

there zero is only not there

Answer 2

Answer: The zeros of the polynomial p(x) = x^2 + 51x - 150 are approximately 2.22 and -53.22.

Step-by-step explanation:

To find the zeros of the quadratic polynomial p(x) = x^2 + 51x - 150, we need to solve for the values of x that make the polynomial equal to zero.

We can use the quadratic formula to find the zeros of the polynomial. The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± √(b^2 - 4ac))/2a

For the polynomial p(x) = x^2 + 51x - 150, we have a = 1, b = 51, and c = -150. Substituting these values into the quadratic formula, we get:

x = (-51 ± √(51^2 - 4(1)(-150)))/2(1)

Simplifying this expression, we get:

x = (-51 ± √(2801))/2

Therefore, the zeros of the polynomial are:

x1 = (-51 + √(2801))/2 ≈ 2.22

x2 = (-51 - √(2801))/2 ≈ -53.22

Thus, the zeros of the polynomial p(x) = x^2 + 51x - 150 are approximately 2.22 and -53.22.

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