How do I solve a polynomial with a missing term?

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In summary: x=\frac{-4\pm\sqrt{8^2-4(1)(55)}}{2(1)}=-4\pm\sqrt{39}i$x^3 - 9x -440$$=x^3 - 8x^2 + 8x^2 - 64x + 55x - 440$$=x^2(x - 8) + 8x(x - 8) + 55(x - 8)$$=(x - 8)(x^2 + 8x + 55)$$=(x - 8)(x^2 + 8x + 16 -16 + 55)$
  • #1
blinky
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0
Hi,

I'm trying to help a high-school sophomore with a math problem, and unfortunately my algebra days are long behind me. Here's the equation:

x^3-9x-440=0

I know x=8, but I don't know how to find it. I'd appreciate some guidance.

Thanks.
 
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  • #2
blinky said:
Hi,

I'm trying to help a high-school sophomore with a math problem, and unfortunately my algebra days are long behind me. Here's the equation:

x^3-9x-440=0

I know x=8, but I don't know how to find it. I'd appreciate some guidance.

Thanks.

use the rational root theorem. if x an integer is a zero it is factor of 440

so look at them 1,2,4,5,8,10,11,20,22,40,44,55,88,110,220,440 ( both plus and minus)

as the sign changes once so there is a positive root
x^3 = 440 + 9x or x^3 > 440 or x > 7
we need to check for values > 7( from the above set)
try 8 and it succeed
so we need not try further
you can divide by x-8 and get a quadratic and solve the same
 
  • #3
kaliprasad said:
use the rational root theorem. if x an integer is a zero it is factor of 440

so look at them 1,2,4,5,8,10,11,20,22,40,44,55,88,110,220,440 ( both plus and minus)

as the sign changes once so there is a positive root
x^3 = 440 + 9x or x^3 > 440 or x > 7
we need to check for values > 7( from the above set)
try 8 and it succeed
so we need not try further
you can divide by x-8 and get a quadratic and solve the same

Thanks very much. I'll tell her.

Did you know just from looking at the equation that it had to be solved that way? If so, how did you know?
 
  • #4
blinky said:
Thanks very much. I'll tell her.

Did you know just from looking at the equation that it had to be solved that way? If so, how did you know?

factoring is the way out. but if you cannot factor obviously, then we can use the rational root theorem to find the factor. This is one of the ways
 
  • #5
kaliprasad said:
factoring is the way out. but if you cannot factor obviously, then we can use the rational root theorem to find the factor. This is one of the ways

Thanks again.
 
  • #6
blinky said:
Hi,

I'm trying to help a high-school sophomore with a math problem, and unfortunately my algebra days are long behind me. Here's the equation:

x^3-9x-440=0

I know x=8, but I don't know how to find it. I'd appreciate some guidance.

Thanks.

Knowing that $x=8$, we could them proceed to rewrite the equation as:

\(\displaystyle x^3-8x^2+8x^2-64x+55x-440=0\)

Factor:

\(\displaystyle x^2(x-8)+8x(x-8)+55(x-8)=0\)

\(\displaystyle (x-8)\left(x^2+8x+55\right)=0\)

To get the other 2 roots, we apply the quadratic formula to the quadratic factor:

\(\displaystyle x=\frac{-8\pm\sqrt{8^2-4(1)(55)}}{2(1)}=-4\pm\sqrt{39}i\)
 
  • #7
$x^3 - 9x -440$
$=x^3 - 8x^2 + 8x^2 - 64x + 55x - 440$
$=x^2(x - 8) + 8x(x - 8) + 55(x - 8)$
$=(x - 8)(x^2 + 8x + 55)$
$=(x - 8)(x^2 + 8x + 16 -16 + 55)$
$=(x - 8)((x + 4)^2 + 39)$
$=(x - 8)(x + 4)^2 - (-39))$
$=(x - 8)(x + 4)^2 - (\sqrt{39}i)^2$
$=(x - 8)(x + 4 - \sqrt{39}i)(x + 4 + \sqrt{39}i)$
 

Related to How do I solve a polynomial with a missing term?

1. What is a polynomial?

A polynomial is a mathematical expression that consists of variables and coefficients, combined using operations such as addition, subtraction, multiplication, and non-negative integer exponents.

2. How do you solve a polynomial?

To solve a polynomial, you need to find the values of the variables that make the expression equal to a specific number, usually zero. This can be done by applying various techniques such as factoring, synthetic division, or the quadratic formula.

3. What is the difference between a linear and quadratic polynomial?

A linear polynomial has a degree of 1, meaning it contains only one variable raised to the first power. A quadratic polynomial has a degree of 2, meaning it contains a variable raised to the second power. Additionally, a quadratic polynomial can have a constant term, while a linear polynomial cannot.

4. Can you solve a polynomial with more than one variable?

Yes, it is possible to solve a polynomial with more than one variable. However, the process becomes more complex as the number of variables increases. In such cases, the polynomial is often solved by using a system of equations.

5. Are there any real-world applications of solving polynomials?

Yes, solving polynomials has many real-world applications, such as in engineering, physics, and economics. For example, it can be used to model and solve problems related to motion, population growth, and financial investments.

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