Given The Polynomial $P(x) = 6x^3 - 10x^2 + 15x - 15$, Calculate The Following Values:1) $P(2$\]2) $P(3$\]3) $P(-4$\]4) $P(-3$\]5) $P(-2$\]6) $P(7$\]7) $P(-1$\]8) $P(-10$\]9)

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In this article, we will explore the process of evaluating polynomials, specifically the given polynomial P(x)=6x3−10x2+15x−15P(x) = 6x^3 - 10x^2 + 15x - 15. We will calculate the values of P(x)P(x) at various points, including x=2,3,−4,−3,−2,7,−1,−10x = 2, 3, -4, -3, -2, 7, -1, -10.

What is a Polynomial?

A polynomial is an expression consisting of variables and coefficients combined using only addition, subtraction, and multiplication. Polynomials can be classified based on the degree of the highest power of the variable. For example, a polynomial of degree 3 is a cubic polynomial, while a polynomial of degree 2 is a quadratic polynomial.

Evaluating Polynomials

To evaluate a polynomial at a given point, we substitute the value of the variable into the polynomial expression. Let's consider the given polynomial P(x)=6x3−10x2+15x−15P(x) = 6x^3 - 10x^2 + 15x - 15. We will evaluate this polynomial at the points x=2,3,−4,−3,−2,7,−1,−10x = 2, 3, -4, -3, -2, 7, -1, -10.

Evaluating P(2)P(2)

To evaluate P(2)P(2), we substitute x=2x = 2 into the polynomial expression:

P(2)=6(2)3−10(2)2+15(2)−15P(2) = 6(2)^3 - 10(2)^2 + 15(2) - 15

P(2)=6(8)−10(4)+30−15P(2) = 6(8) - 10(4) + 30 - 15

P(2)=48−40+30−15P(2) = 48 - 40 + 30 - 15

P(2)=23P(2) = 23

Evaluating P(3)P(3)

To evaluate P(3)P(3), we substitute x=3x = 3 into the polynomial expression:

P(3)=6(3)3−10(3)2+15(3)−15P(3) = 6(3)^3 - 10(3)^2 + 15(3) - 15

P(3)=6(27)−10(9)+45−15P(3) = 6(27) - 10(9) + 45 - 15

P(3)=162−90+45−15P(3) = 162 - 90 + 45 - 15

P(3)=102P(3) = 102

Evaluating P(−4)P(-4)

To evaluate P(−4)P(-4), we substitute x=−4x = -4 into the polynomial expression:

P(−4)=6(−4)3−10(−4)2+15(−4)−15P(-4) = 6(-4)^3 - 10(-4)^2 + 15(-4) - 15

P(−4)=6(−64)−10(16)−60−15P(-4) = 6(-64) - 10(16) - 60 - 15

P(−4)=−384−160−60−15P(-4) = -384 - 160 - 60 - 15

P(−4)=−619P(-4) = -619

Evaluating P(−3)P(-3)

To evaluate P(−3)P(-3), we substitute x=−3x = -3 into the polynomial expression:

P(−3)=6(−3)3−10(−3)2+15(−3)−15P(-3) = 6(-3)^3 - 10(-3)^2 + 15(-3) - 15

P(−3)=6(−27)−10(9)−45−15P(-3) = 6(-27) - 10(9) - 45 - 15

P(−3)=−162−90−45−15P(-3) = -162 - 90 - 45 - 15

P(−3)=−312P(-3) = -312

Evaluating P(−2)P(-2)

To evaluate P(−2)P(-2), we substitute x=−2x = -2 into the polynomial expression:

P(−2)=6(−2)3−10(−2)2+15(−2)−15P(-2) = 6(-2)^3 - 10(-2)^2 + 15(-2) - 15

P(−2)=6(−8)−10(4)−30−15P(-2) = 6(-8) - 10(4) - 30 - 15

P(−2)=−48−40−30−15P(-2) = -48 - 40 - 30 - 15

P(−2)=−133P(-2) = -133

Evaluating P(7)P(7)

To evaluate P(7)P(7), we substitute x=7x = 7 into the polynomial expression:

P(7)=6(7)3−10(7)2+15(7)−15P(7) = 6(7)^3 - 10(7)^2 + 15(7) - 15

P(7)=6(343)−10(49)+105−15P(7) = 6(343) - 10(49) + 105 - 15

P(7)=2058−490+105−15P(7) = 2058 - 490 + 105 - 15

P(7)=1658P(7) = 1658

Evaluating P(−1)P(-1)

To evaluate P(−1)P(-1), we substitute x=−1x = -1 into the polynomial expression:

P(−1)=6(−1)3−10(−1)2+15(−1)−15P(-1) = 6(-1)^3 - 10(-1)^2 + 15(-1) - 15

P(−1)=6(−1)−10(1)−15−15P(-1) = 6(-1) - 10(1) - 15 - 15

P(−1)=−6−10−15−15P(-1) = -6 - 10 - 15 - 15

P(−1)=−46P(-1) = -46

Evaluating P(−10)P(-10)

To evaluate P(−10)P(-10), we substitute x=−10x = -10 into the polynomial expression:

P(−10)=6(−10)3−10(−10)2+15(−10)−15P(-10) = 6(-10)^3 - 10(-10)^2 + 15(-10) - 15

P(−10)=6(−1000)−10(100)−150−15P(-10) = 6(-1000) - 10(100) - 150 - 15

P(−10)=−6000−1000−150−15P(-10) = -6000 - 1000 - 150 - 15

P(−10)=−7165P(-10) = -7165

Conclusion

In this article, we evaluated the polynomial P(x)=6x3−10x2+15x−15P(x) = 6x^3 - 10x^2 + 15x - 15 at various points, including x=2,3,−4,−3,−2,7,−1,−10x = 2, 3, -4, -3, -2, 7, -1, -10. We used the process of substitution to find the values of P(x)P(x) at each point. The results are as follows:

  • P(2)=23P(2) = 23
  • P(3)=102P(3) = 102
  • P(−4)=−619P(-4) = -619
  • P(−3)=−312P(-3) = -312
  • P(−2)=−133P(-2) = -133
  • P(7)=1658P(7) = 1658
  • P(−1)=−46P(-1) = -46
  • P(−10)=−7165P(-10) = -7165

In our previous article, we explored the process of evaluating polynomials, specifically the given polynomial P(x)=6x3−10x2+15x−15P(x) = 6x^3 - 10x^2 + 15x - 15. We calculated the values of P(x)P(x) at various points, including x=2,3,−4,−3,−2,7,−1,−10x = 2, 3, -4, -3, -2, 7, -1, -10. In this article, we will address some common questions and concerns related to evaluating polynomials.

Q: What is the difference between evaluating a polynomial and solving a polynomial equation?

A: Evaluating a polynomial involves substituting a specific value of the variable into the polynomial expression to find the corresponding value of the polynomial. Solving a polynomial equation, on the other hand, involves finding the values of the variable that make the polynomial equal to zero.

Q: How do I know which value to substitute into the polynomial expression?

A: When evaluating a polynomial, you need to substitute the specific value of the variable that you are interested in. For example, if you want to find the value of the polynomial at x=2x = 2, you would substitute x=2x = 2 into the polynomial expression.

Q: Can I use a calculator to evaluate a polynomial?

A: Yes, you can use a calculator to evaluate a polynomial. In fact, calculators are often the most efficient way to evaluate polynomials, especially for large or complex polynomials.

Q: What if I make a mistake when evaluating a polynomial?

A: If you make a mistake when evaluating a polynomial, you may get an incorrect answer. To avoid this, it's essential to double-check your work and use a calculator or other tools to verify your results.

Q: Can I use the same method to evaluate a polynomial with multiple variables?

A: No, the method we discussed in this article is only applicable to polynomials with a single variable. If you have a polynomial with multiple variables, you will need to use a different method to evaluate it.

Q: How do I know if a polynomial is easy or hard to evaluate?

A: The difficulty of evaluating a polynomial depends on the complexity of the polynomial expression and the value of the variable. In general, polynomials with higher degrees or more complex expressions are harder to evaluate.

Q: Can I use the same method to evaluate a polynomial with negative coefficients?

A: Yes, the method we discussed in this article is applicable to polynomials with negative coefficients. However, you may need to use a calculator or other tools to handle the negative coefficients.

Q: What if I need to evaluate a polynomial at multiple points?

A: If you need to evaluate a polynomial at multiple points, you can use a calculator or other tools to simplify the process. Alternatively, you can use a programming language or software to automate the evaluation process.

Conclusion

In this article, we addressed some common questions and concerns related to evaluating polynomials. We discussed the difference between evaluating a polynomial and solving a polynomial equation, how to choose the value to substitute into the polynomial expression, and how to use a calculator to evaluate a polynomial. We also covered some common mistakes to avoid and how to evaluate polynomials with multiple variables or negative coefficients. By following these guidelines, you can become more confident and proficient in evaluating polynomials.

Additional Resources

Practice Problems

  • Evaluate the polynomial P(x)=2x2+3x−1P(x) = 2x^2 + 3x - 1 at x=4x = 4.
  • Evaluate the polynomial P(x)=x3−2x2+x+1P(x) = x^3 - 2x^2 + x + 1 at x=−2x = -2.
  • Evaluate the polynomial P(x)=2x2−3x+1P(x) = 2x^2 - 3x + 1 at x=1x = 1.

Answer Key

  • P(4)=2(4)2+3(4)−1=32+12−1=43P(4) = 2(4)^2 + 3(4) - 1 = 32 + 12 - 1 = 43
  • P(−2)=(−2)3−2(−2)2+(−2)+1=−8−8−2+1=−17P(-2) = (-2)^3 - 2(-2)^2 + (-2) + 1 = -8 - 8 - 2 + 1 = -17
  • P(1)=2(1)2−3(1)+1=2−3+1=0P(1) = 2(1)^2 - 3(1) + 1 = 2 - 3 + 1 = 0