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for the equation: -2hÂ² - 28h - 98 = 0 can be factored as follows: -2(hÂ² + 14h + 49) = 0;

to solve, the expression in parenthesis must equal zero: (hÂ² + 14h + 49) = 0. We are looking for values, which when added equal 14, and when multiplied equal 49.

(h + a)(h + b) = 0, where we want a+b = 14, and ab = 49. The answer is a = 7 and b = 7, so we have (h + 7)(h + 7) = 0, and h = -7,-7 [double root].

Q: How to solve -2h2-28h-98 equals 0 by factoring polynomials?

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(3x+4)(3x-4)=0 x=±4/3

In real life you will probably never divide polynomials, but you need to know how to solve homework and exam problems.

To solve all sorts of problems. Any equation can be written in the form: (some expression) = 0 Simply by putting everything to the left. It turns out that polynomials are especially easy to solve if you put them in that form - because then you can solve them simply by factoring them. In other cases, for other functions, it might be more of a convention to put them in that form.

In this case, the left part of the equation can easily be factored. If you are not familiar with factoring polynomials, use the quadratic formula.================================================v2 + 6v + 8 = 0(v + 4) (v + 2) = 0v = -4andv = -2

If there is no common factor then the polynomial cannot be factorised. If there is no common factor then the polynomial cannot be factorised. If there is no common factor then the polynomial cannot be factorised. If there is no common factor then the polynomial cannot be factorised.

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2x2y2+5=0 how to solve this

(3x+4)(3x-4)=0 x=±4/3

In real life you will probably never divide polynomials, but you need to know how to solve homework and exam problems.

To solve all sorts of problems. Any equation can be written in the form: (some expression) = 0 Simply by putting everything to the left. It turns out that polynomials are especially easy to solve if you put them in that form - because then you can solve them simply by factoring them. In other cases, for other functions, it might be more of a convention to put them in that form.

By factoring. q2 + 16q = 0 q (q + 16) = 0 Now, either q = 0, or q + 16 = 0. Solve those two equations to get the solution.

x2 - x = 12 Rewrite as x2 - x - 12 = 0 Now you can solve by factoring as you do most polynomials. (x - ?)(x + ?) Find the factors of -12 that also add up to -1 In this case, -4 and 3 (x - 4)(x + 3) = 0 Solve individually because anything multiplied by zero is zero x - 4 = 0 x = 4 x + 3 = 0 x = -3 So your answers are -3,4. You can check by popping them into the original equation.

When the co-efficient of x2 is 1

In this case, the left part of the equation can easily be factored. If you are not familiar with factoring polynomials, use the quadratic formula.================================================v2 + 6v + 8 = 0(v + 4) (v + 2) = 0v = -4andv = -2

If there is no common factor then the polynomial cannot be factorised. If there is no common factor then the polynomial cannot be factorised. If there is no common factor then the polynomial cannot be factorised. If there is no common factor then the polynomial cannot be factorised.

Solve by factoring. Solve by taking the square root of both sides.

y=b+x+x^2 This is a quadratic equation. The graph is a parabola. The quadratic equation formula or factoring can be used to solve this.