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jeffmarina
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Homework Statement
x^2(4x + 13) = 9
Homework Equations
The Attempt at a Solution
Answers are that x = -3, -1 and 3/4. But how can one compute this other than trial and error?
Hi, gabba just want to make sure I understand your method here. Is it possible that this may not work for some polynomials since we may end up with a cubic equation of a variable say x_1 ? Which of course leads us back to square one.gabbagabbahey said:Okay good, you can get rid of the 0x term though :)
The next step is to recognize that since this is a third degree polynomial, there will be three roots; let's call them [itex]x_1[/itex], [itex]x_2[/itex], and [itex]x_3[/itex]. This means that [itex]4(x-x_1)(x-x_2)(x-x_3)=0=4x^3+13x^2-9[/itex]. Expand the expression on the left. What do you get?
Defennder said:Hi, gabba just want to make sure I understand your method here. Is it possible that this may not work for some polynomials since we may end up with a cubic equation of a variable say x_1 ? Which of course leads us back to square one.
To solve a third degree polynomial, also known as a cubic equation, you can use the cubic formula or factor it into a product of linear and quadratic expressions.
The cubic formula is a formula used to solve cubic equations of the form ax³ + bx² + cx + d = 0. It involves finding the roots of the equation using complex numbers.
To factor a third degree polynomial, you can use the rational root theorem to find possible rational roots, and then use synthetic division or long division to factor it into a product of linear and quadratic expressions.
No, the quadratic formula can only be used to solve quadratic equations. For third degree polynomials, you will need to use the cubic formula or factor it into linear and quadratic expressions.
A third degree polynomial can have up to three solutions, as it is a cubic equation. However, it is possible for the equation to have less than three solutions or no real solutions at all.