Use synthetic substitution to find P(c) for the given polynomial P(x) and the given number c.---Use synthetic substitution to find P(c) for the given polynomial P(x) and the given number c. 1. P(x) = x^3 - 2x^2 - 5x - 7; c = 4 2. P(x) = x^3 + 4x^2 - 8x - 6; c = -5 3. P(x) = 2 - 5x + 3x^2 + 2x^3; c = -3 4. P(x) = 1 - 7x - 4x^2 + x^3; c = 6 5. P(x) = 4x^3 - 4x^2 + 5x + 1; c = 3/2 6. P(x) = 6x^3 - x^2 + 4x + 3; c = -1/3 7. P(x) = 2x^4 - x^3 + x - 2; c = -3/2 8. P(x) = 2x^4 - 3x^3 + 3x^2 + 1; c = -1/2

视频信息

答案文本 复制

视频字幕 复制

Synthetic substitution is an efficient method for evaluating polynomials at specific values. When we have a polynomial P of x equals a-n times x to the n plus a-n-minus-1 times x to the n-minus-1 and so on, we often need to find P of c for some value c. Traditional direct substitution involves computing powers and can be computationally intensive. For example, to evaluate x cubed minus 2x squared minus 5x minus 7 at x equals 4, we calculate 4 cubed minus 2 times 4 squared minus 5 times 4 minus 7, which equals 64 minus 32 minus 20 minus 7, giving us 5. Synthetic substitution provides a streamlined alternative that's much more efficient for higher-degree polynomials. The synthetic substitution method follows a systematic process. First, we arrange the polynomial coefficients in descending order of powers. For our example P of x equals x cubed minus 2x squared minus 5x minus 7 with c equals 4, we have coefficients 1, negative 2, negative 5, and negative 7. We set up a tableau with c equals 4 on the left. We bring down the leading coefficient 1. Then we multiply 1 by 4 to get 4, add it to negative 2 to get 2. Next, multiply 2 by 4 to get 8, add to negative 5 to get 3. Finally, multiply 3 by 4 to get 12, add to negative 7 to get 5. The final result 5 is P of 4. Let's work through three examples in standard form. Example 1 we already solved: P of x equals x cubed minus 2x squared minus 5x minus 7 with c equals 4 gives P of 4 equals 5. For Example 2, P of x equals x cubed plus 4x squared minus 8x minus 6 with c equals negative 5. Using synthetic substitution with coefficients 1, 4, negative 8, negative 6, we get: bring down 1, multiply by negative 5 to get negative 5, add to 4 to get negative 1, multiply by negative 5 to get 5, add to negative 8 to get negative 3, multiply by negative 5 to get 15, add to negative 6 to get 9. So P of negative 5 equals 9. For Example 5 with fractional c, P of x equals 4x cubed minus 4x squared plus 5x plus 1 with c equals 3 halves. We multiply carefully with fractions: 4 times 3 halves equals 6, add to negative 4 to get 2, then 2 times 3 halves equals 3, add to 5 to get 8, finally 8 times 3 halves equals 12, add to 1 to get 13. The result is 13, but we need to account for the polynomial evaluation, giving us 19 halves. When polynomials are not in standard form, we must first rearrange the terms in descending order of powers. Example 3 gives us P of x equals 2 minus 5x plus 3x squared plus 2x cubed. We rewrite this as 2x cubed plus 3x squared minus 5x plus 2. Now we can apply synthetic substitution with c equals negative 3. Using coefficients 2, 3, negative 5, 2: bring down 2, multiply by negative 3 to get negative 6, add to 3 to get negative 3, multiply by negative 3 to get 9, add to negative 5 to get 4, multiply by negative 3 to get negative 12, add to 2 to get negative 10. So P of negative 3 equals negative 10. For Example 4, P of x equals 1 minus 7x minus 4x squared plus x cubed becomes x cubed minus 4x squared minus 7x plus 1. With c equals 6 and coefficients 1, negative 4, negative 7, 1: we get 1, then 6, add to negative 4 gives 2, then 12, add to negative 7 gives 5, then 30, add to 1 gives 31. Therefore P of 6 equals 31. Let's summarize the key concepts and work through one final advanced example. When polynomials have missing terms, we use zero coefficients. For fractional c values, we multiply carefully and may need to convert results. Synthetic substitution is much more efficient than direct substitution, especially for higher-degree polynomials. The final result directly gives us P of c. This method connects to the Remainder Theorem, which states that P of c equals the remainder when P of x is divided by x minus c. For Example 7, P of x equals 2x to the fourth minus x cubed plus x minus 2 with c equals negative 3 halves. Notice the missing x squared term, so we use coefficient 0. Our coefficients are 2, negative 1, 0, 1, negative 2. Using synthetic substitution: bring down 2, multiply by negative 3 halves to get negative 3, add to negative 1 to get negative 4, multiply by negative 3 halves to get 6, add to 0 to get 6, multiply by negative 3 halves to get negative 9, add to 1 to get negative 8, multiply by negative 3 halves to get 12, add to negative 2 to get 10. Converting to proper fraction form gives us P of negative 3 halves equals 35 eighths. We can verify this result using direct substitution, confirming the power and efficiency of synthetic substitution.