# Extra Questions For Class 9 Maths Polynomials | Q5

###### Chapter 2 | Factorise Quadratic | Remainder & Factor Theorem

Question 5: Without actual division prove that (x2 - x - 2) divides 2x4 + x3 - 5x2 - 8x - 4

## NCERT Solution to Class 9 Maths

### Explanatory Answer | Chapter 2 Extra Question 5

#### Approach to Solve this Polynomials Question

Let p(x) = 2x4 + x3 - 5x2 - 8x - 4

1. Step 1: Factorize the quadratic expression (x2 - x - 2) - the divisor - by splitting the middle term. Let it factorize as (x - a)(x - b)
2. Step 2: Put the zeroes 'a' and 'b' of factorizing the quadratic polynomial in p(x) and find p(a) and p(b)
3. Step 3: If p(a) and p(b) are both equal to zero, then 'a' and 'b' are zeroes of p(x)
4. Step 4: If 'a' and 'b' are zeroes of p(x), by Factor Theorem, (x - a) and (x - b) are factors of p(x)
5. 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 (x2 - x - 2)

### Stepwise Solution

Let p(x) = 2x4 + x3 - 5x2 - 8x - 4

#### Step 1: Factorize the quadratic polynomial (x2 - x - 2) by splitting the middle term

x2 - x - 2 = x2 - 2x + x - 2
= x(x - 2) +1(x - 2)
= (x - 2)(x + 1)
Hence, zeroes of x2 - x - 2 are 2 and (-1)

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

p(2) = 2(24) + 23 - 5(22) - 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) = x2 - x - 2 is a factor of p(x).

Or, x2 - x - 2 will divide p(x) without leaving a remainder.

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