Apply the Factor Theorem.

Question

Apply the Factor Theorem.

Asked on March 28, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-03-28T10:05:57.908Z

6. Determine whether $x-1$ is a factor of the polynomial $P(x) = x^3 - 4x^2 + x + 6$. Step 1: Apply the Factor Theorem. According to the Factor Theorem, $(x-1)$ is a factor of $P(x)$ if and only if $P(1) = 0$. Step 2: Evaluate $P(1)$. $$ P(1) = (1)^3 - 4(1)^2 + (1) + 6 $$ $$ P(1) = 1 - 4(1) + 1 + 6 $$ $$ P(1) = 1 - 4 + 1 + 6 $$ $$ P(1) = -3 + 1 + 6 $$ $$ P(1) = -2 + 6 $$ $$ P(1) = 4 $$ Step 3: Conclude based on the result. Since $P(1) = 4 \neq 0$, $(x-1)$ is not a factor of $P(x)$. The answer is $\boxed{\text{No, } x-1 \text{ is not a factor}}$. 7. Given that $x+2$ is a factor of the polynomial $P(x) = x^3 + x^2 - 4x - 4$, find the complete factorization of $P(x)$. Step 1: Use synthetic division to divide $P(x)$ by $(x+2)$. Since $(x+2)$ is a factor, we divide by $x = -2$. ` -2 | 1 1 -4 -4 | -2 2 4 ------------------ 1 -1 -2 0 ` The quotient is $x^2 - x - 2$ and the remainder is 0. Step 2: Factor the quadratic quotient. The quadratic quotient is $x^2 - x - 2$. We need two numbers that multiply to -2 and add to -1. These numbers are -2 and 1. $$ x^2 - x - 2 = (x - 2)(x + 1) $$ Step 3: Write the complete factorization of $P(x)$. $$ P(x) = (x+2)(x^2 - x - 2) $$ $$ P(x) = (x+2)(x-2)(x+1) $$ The complete factorization is $\boxed{P(x) = (x+2)(x-2)(x+1)}$. 8. Explain why the following expression is not a polynomial: $f(x) = \frac{3}{x^2} + x$. A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. The term $\frac{3}{x^2}$ can be written as $3x^{-2}$. Since the exponent of $x$ in this term is $-2$, which is a negative integer, the expression $f(x)$ does not meet the definition of a polynomial. 9. Find the remainder when $P(x) = 3x^4 - 2x^3 + 5x - 7$ is divided by $x+1$. Step 1: Apply the Remainder Theorem. According to the Remainder Theorem, the remainder when $P(x)$ is divided by $(x+1)$ is $P(-1)$. Step 2: Evaluate $P(-1)$. $$ P(-1) = 3(-1)^4 - 2(-1)^3 + 5(-1) - 7 $$ $$ P(-1) = 3(1) - 2(-1) - 5 - 7 $$ $$ P(-1) = 3 + 2 - 5 - 7 $$ $$ P(-1) = 5 - 5 - 7 $$ $$ P(-1) = 0 - 7 $$ $$ P(-1) = -7 $$ The remainder is $\boxed{-7}$. 10. Let $P(x) = x^3 + ax^2 + bx + c$ be a polynomial with real coefficients. Suppose that $P(1)=6$, $P(2)=15$, $P(3)=28$. Find the values of $a, b$, and $c$. Hence determine the polynomial $P(x)$. Step 1: Set up a system of equations using the given conditions. For $P(1)=6$: $$ (1)^3 + a(1)^2 + b(1) + c = 6 $$ $$ 1 + a + b + c = 6 $$ (1) $a + b + c = 5$ For $P(2)=15$: $$ (2)^3 + a(2)^2 + b(2) + c = 15 $$ $$ 8 + 4a + 2b + c = 15 $$ (2) $4a + 2b + c = 7$ For $P(3)=28$: $$ (3)^3 + a(3)^2 + b(3) + c = 28 $$ $$ 27 + 9a + 3b + c = 28 $$ (3) $9a + 3b + c = 1$ Step 2: Solve the system of linear equations. Subtract (1) from (2): $$ (4a + 2b + c) - (a + b + c) = 7 - 5 $$ (4) $3a + b = 2$ Subtract (2) from (3): $$ (9a + 3b + c) - (4a + 2b + c) = 1 - 7 $$ (5) $5a + b = -6$ Subtract (4) from (5): $$ (5a + b) - (3a + b) = -6 - 2 $$ $$ 2a = -8 $$ $$ a = -4 $$ Substitute $a = -4$ into (4): $$ 3(-4) + b = 2 $$ $$ -12 + b = 2 $$ $$ b = 14 $$ Substitute $a = -4$ and $b = 14$ into (1): $$ -4 + 14 + c = 5 $$ $$ 10 + c = 5 $$ $$ c = -5 $$ Step 3: Determine the polynomial $P(x)$. Substitute the values $a = -4$, $b = 14$, and $c = -5$ into $P(x) = x^3 + ax^2 + bx + c$. $$ P(x) = x^3 - 4x^2 + 14x - 5 $$ The values are $\boxed{a = -4, b = 14, c = -5}$. The polynomial is $\boxed{P(x) = x^3 - 4x^2 + 14x - 5}$.

Accepted Answer · 2026-03-28T10:05:57.908Z

6. Determine whether $x-1$ is a factor of the polynomial $P(x) = x^3 - 4x^2 + x + 6$. Step 1: Apply the Factor Theorem. According to the Factor Theorem, $(x-1)$ is a factor of $P(x)$ if and only if $P(1) = 0$. Step 2: Evaluate $P(1)$. $$ P(1) = (1)^3 - 4(1)^2 + (1) + 6 $$ $$ P(1) = 1 - 4(1) + 1 + 6 $$ $$ P(1) = 1 - 4 + 1 + 6 $$ $$ P(1) = -3 + 1 + 6 $$ $$ P(1) = -2 + 6 $$ $$ P(1) = 4 $$ Step 3: Conclude based on the result. Since $P(1) = 4 \neq 0$, $(x-1)$ is not a factor of $P(x)$. The answer is $\boxed{\text{No, } x-1 \text{ is not a factor}}$. 7. Given that $x+2$ is a factor of the polynomial $P(x) = x^3 + x^2 - 4x - 4$, find the complete factorization of $P(x)$. Step 1: Use synthetic division to divide $P(x)$ by $(x+2)$. Since $(x+2)$ is a factor, we divide by $x = -2$. ` -2 | 1 1 -4 -4 | -2 2 4 ------------------ 1 -1 -2 0 ` The quotient is $x^2 - x - 2$ and the remainder is 0. Step 2: Factor the quadratic quotient. The quadratic quotient is $x^2 - x - 2$. We need two numbers that multiply to -2 and add to -1. These numbers are -2 and 1. $$ x^2 - x - 2 = (x - 2)(x + 1) $$ Step 3: Write the complete factorization of $P(x)$. $$ P(x) = (x+2)(x^2 - x - 2) $$ $$ P(x) = (x+2)(x-2)(x+1) $$ The complete factorization is $\boxed{P(x) = (x+2)(x-2)(x+1)}$. 8. Explain why the following expression is not a polynomial: $f(x) = \frac{3}{x^2} + x$. A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. The term $\frac{3}{x^2}$ can be written as $3x^{-2}$. Since the exponent of $x$ in this term is $-2$, which is a negative integer, the expression $f(x)$ does not meet the definition of a polynomial. 9. Find the remainder when $P(x) = 3x^4 - 2x^3 + 5x - 7$ is divided by $x+1$. Step 1: Apply the Remainder Theorem. According to the Remainder Theorem, the remainder when $P(x)$ is divided by $(x+1)$ is $P(-1)$. Step 2: Evaluate $P(-1)$. $$ P(-1) = 3(-1)^4 - 2(-1)^3 + 5(-1) - 7 $$ $$ P(-1) = 3(1) - 2(-1) - 5 - 7 $$ $$ P(-1) = 3 + 2 - 5 - 7 $$ $$ P(-1) = 5 - 5 - 7 $$ $$ P(-1) = 0 - 7 $$ $$ P(-1) = -7 $$ The remainder is $\boxed{-7}$. 10. Let $P(x) = x^3 + ax^2 + bx + c$ be a polynomial with real coefficients. Suppose that $P(1)=6$, $P(2)=15$, $P(3)=28$. Find the values of $a, b$, and $c$. Hence determine the polynomial $P(x)$. Step 1: Set up a system of equations using the given conditions. For $P(1)=6$: $$ (1)^3 + a(1)^2 + b(1) + c = 6 $$ $$ 1 + a + b + c = 6 $$ (1) $a + b + c = 5$ For $P(2)=15$: $$ (2)^3 + a(2)^2 + b(2) + c = 15 $$ $$ 8 + 4a + 2b + c = 15 $$ (2) $4a + 2b + c = 7$ For $P(3)=28$: $$ (3)^3 + a(3)^2 + b(3) + c = 28 $$ $$ 27 + 9a + 3b + c = 28 $$ (3) $9a + 3b + c = 1$ Step 2: Solve the system of linear equations. Subtract (1) from (2): $$ (4a + 2b + c) - (a + b + c) = 7 - 5 $$ (4) $3a + b = 2$ Subtract (2) from (3): $$ (9a + 3b + c) - (4a + 2b + c) = 1 - 7 $$ (5) $5a + b = -6$ Subtract (4) from (5): $$ (5a + b) - (3a + b) = -6 - 2 $$ $$ 2a = -8 $$ $$ a = -4 $$ Substitute $a = -4$ into (4): $$ 3(-4) + b = 2 $$ $$ -12 + b = 2 $$ $$ b = 14 $$ Substitute $a = -4$ and $b = 14$ into (1): $$ -4 + 14 + c = 5 $$ $$ 10 + c = 5 $$ $$ c = -5 $$ Step 3: Determine the polynomial $P(x)$. Substitute the values $a = -4$, $b = 14$, and $c = -5$ into $P(x) = x^3 + ax^2 + bx + c$. $$ P(x) = x^3 - 4x^2 + 14x - 5 $$ The values are $\boxed{a = -4, b = 14, c = -5}$. The polynomial is $\boxed{P(x) = x^3 - 4x^2 + 14x - 5}$.