Given polynomial P(x) = x³ + x² − 6x − 4. a) Show that (x−2) is a factor of P(x). b) Factorise P(x)…

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Given polynomial P(x) = x³ + x² − 6x − 4. a) Show that (x−2) is a factor of P(x). b) Factorise P(x) completely. c) Hence, solve the equation P(x) = 0.

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The given polynomial is f(x) = x³ + x² − 4x − 4. a) Show that (x−2) is a factor of f(x). Step 1: According to the Factor Theorem, if (x−a) is a factor of a polynomial f(x), then f(a) = 0. In this case, a=2. Step 2: Substitute x=2 into the polynomial f(x). f(2) = (2)³ + (2)² − 4(2) − 4 Step 3: Evaluate the expression. f(2) = 8 + 4 − 8 − 4; f(2) = 12 − 12; f(2) = 0; Since f(2) = 0, (x−2) is indeed a factor of f(x). b) Factorize f(x) completely. Step 1: Since (x−2) is a factor, we can perform polynomial division or synthetic division to find the other factor. Using synthetic division with root x=2: 2 | 1 1 -4 -4 | 2 6 4 ----------------- 1 3 2 0 The quotient is x² + 3x + 2. Step 2: Write f(x) as the product of (x−2) and the quadratic factor. f(x) = (x−2)(x² + 3x + 2) Step 3: Factorize the quadratic factor x² + 3x + 2. We look for two numbers that multiply to 2 and add to 3. These numbers are 1 and 2. x² + 3x + 2 = (x+1)(x+2) Step 4: Write the completely factorized form of f(x). f(x) = (x−2)(x+1)(x+2) The completely factorized form is (x−2)(x+1)(x+2). c) Hence, solve the equation f(x) = 0. Step 1: Set the completely factorized form of f(x) equal to 0. (x−2)(x+1)(x+2) = 0 Step 2: For the product of factors to be zero, at least one of the factors must be zero. Set each factor equal to zero and solve for x. x−2 = 0 ⇒ x = 2; x+1 = 0 ⇒ x = −1; x+2 = 0 ⇒ x = −2; The solutions to the equation f(x)=0 are x = 2, −1, −2.

Given polynomial P(x) = x³ + x² − 6x − 4. a) Show that (x-2) is a factor of P(x). b) Factorise P(x) completely. c) Hence, solve the equation P(x) = 0.

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Step-by-step solution

Part a· 3 steps

According to the Factor Theorem, if $(x-a)$ is a factor of a polynomial $f(x)$ , then $f(a) = 0$ .

Substitute $x=2$ into the polynomial $f(x)$ .

Evaluate the expression.

Part b· 4 steps(x-2)(x+1)(x+2)

Since $(x-2)$ is a factor, we can perform polynomial division or synthetic division to find the other factor.

Write $f(x)$ as the product of $(x-2)$ and the quadratic factor.

Factorize the quadratic factor $x^2 + 3x + 2$ .

Write the completely factorized form of $f(x)$ .

Part c· 2 steps
$x = 2, -1, -2$

Set the completely factorized form of $f(x)$ equal to 0.

For the product of factors to be zero, at least one of the factors must be zero.

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Given polynomial P(x) = x³ + x² − 6x − 4. a) Show that (x-2) is a factor of P(x). b) Factorise P(x) completely. c) Hence, solve the equation P(x) = 0.

Answers

Step-by-step solution

Part a· 3 steps

According to the Factor Theorem, if (x−a)(x-a)(x−a) is a factor of a polynomial f(x)f(x)f(x), then f(a)=0f(a) = 0f(a)=0.

Substitute x=2x=2x=2 into the polynomial f(x)f(x)f(x).