Question

For the given function, determine consecutive values of x between which each real zero is located. f(x) = –11x^4 – 5x^3 – 9x^2 + 12x + 10

Answer 1

Answer:

• [-1, 0]
• [0, 1]

Step-by-step explanation:

Descartes' rule of signs tells you this function, with its signs {- - - + +}, having one sign change, will have one positive real root.

When odd-degree terms have their signs changed, the signs {- + - - +} have three changes, so there will be 1 or 3 negative real roots.

The constant term (10) tells us the y-intercept is positive. The sum of coefficients is -11 -5 -9 +12 +10 = -3, so f(1) < 0 and there is a root between 0 and 1.

When odd-degree coefficients change sign, the sum becomes -11 +5 -9 -12+10 = -17, so there is a root between -1 and 0.

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Synthetic division by (x+1) gives a quotient of -11x^3 +6x^2 -15x +27 -17/(x+1), which has alternating signs, indicating -1 is a lower bound on real roots.

Real roots are located in the intervals [-1, 0] and [0, 1].

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The remaining roots are complex. All roots are irrational.

The attached graph confirms that roots are in the intervals listed here. Newton's method iteration is used to refine these to calculator precision. Dividing them from f(x) gives a quadratic with irrational coefficients and complex roots.