Question 5: Without actual division prove that (x^{2} - x - 2) divides 2x^{4} + x^{3} - 5x^{2} - 8x - 4

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### Explanatory Answer | Chapter 2 Extra Question 5

#### Approach to Solve this Polynomials Question

Let p(x) = 2x^{4} + x^{3} - 5x^{2} - 8x - 4

**Step 1**: Factorize the quadratic expression (x^{2} - x - 2) - the divisor - by splitting the middle term. Let it factorize as (x - a)(x - b) **Step 2**: Put the zeroes 'a' and 'b' of factorizing the quadratic polynomial in p(x) and find p(a) and p(b) **Step 3**: If p(a) and p(b) are both equal to zero, then 'a' and 'b' are zeroes of p(x) **Step 4**: If 'a' and 'b' are zeroes of p(x), by Factor Theorem, (x - a) and (x - b) are factors of p(x) **Step 5**: If (x - a) and (x - b) are factors of p(x), then (x - a)(x - b) will be a factor of p(x). Or, p(x) is divisible by (x^{2} - x - 2)

### Stepwise Solution

Let p(x) = 2x^{4} + x^{3} - 5x^{2} - 8x - 4

#### Step 1: Factorize the quadratic polynomial (x^{2} - x - 2) by splitting the middle term

x^{2} - x - 2 = x^{2} - 2x + x - 2

= x(x - 2) +1(x - 2)

= (x - 2)(x + 1)

Hence, zeroes of x^{2} - x - 2 are 2 and (-1)

#### Step 2: Find p(2) and p(-1)

p(2) = 2(2^{4}) + 2^{3} - 5(2^{2}) - 8(2) - 4

= 32 + 8 - 20 - 16 - 4

**0**

So, 2 is a zero of p(x)

p(-1) = 2(-1)^{4} + (-1)^{3} - 5(-1)^{2} - 8(-1) - 4

2 - 1 - 5 + 8 - 4

**0**

So, (-1) is a zero of p(x)

#### Step 3: Zeroes to Factors

Because 2 is a zero of p(x), (x - 2) is a factor of p(x).

Similarly, because (-1) is a zero of p(x), (x + 1) is a factor of p(x).

#### Step 4: The product of factors is a factor

Because (x - 2) and (x + 1) are factors of p(x), (x - 2)(x + 1) is a factor of p(x)

(x - 2)(x + 1) = x^{2} - x - 2 is a factor of p(x).

Or, x^{2} - x - 2 will divide p(x) without leaving a remainder.