Factor Theorem

1. Let p(x) be a polynomial and a be any real number. If p(a) = 0, then (x–a) is a factor of p(x).

2. If (x–a) is a factor of p(x), then p(a) = 0

**EXERCISE 2.4 (NCERT Solution)**

**Q1: Determine which of the following polynomials has (x + 1) a factor :**

**(i) x**

^{3}+ x^{2}+ x + 1**(ii) x**

^{4}+ x^{3}+ x^{2}+ x + 1**(iii) x**

^{4}+ 3x^{3}+ 3x^{2}+ x + 1**(iv) x**

^{3}– x^{2}– (2 + √2)x + √2
Answer: If (x+1) is a factor of a polynomial p(x), then p(-1) must be equal to zero.

(i) Let p(x) = x

^{3}+ x^{2}+ x + 1
p(-1) = (-1)

^{3}+ (-1)^{2}+ (-1) + 1 = -1 + 1 -1 + 1 = 0
∴ (x-1) is a factor of polynomial x

^{3}+ x^{2}+ x + 1
(ii) Let p(x) = x

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

^{4}+ (-1)^{3}+ (-1)^{2}+ (-1) + 1 = 1 - 1 + 1 -1 + 1 = 1 ≠ 0
∴ (x-1) is not a factor of this polynomial.

(iii) Let p(x) = x

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

^{4}+ 3(-1)^{3}+ 3(-1)^{2}+ (-1) + 1
= 1 + 3(-1) + 3(1) -1 + 1 = 1 - 3 + 3 - 1 = 1 ≠ 0

∴ (x-1) is not a factor of this polynomial.

(iv) Let p(x) = x

^{3}– x^{2}– (2 + √2)x + √2
p(-1) = (-1)

^{3}– (-1)^{2}– (2 + √2)(-1) + √2
= -1 - 1 + 2 + √2 + √2 = 2√2 ≠ 0

∴ (x-1) is not a factor of this polynomial.

**Q2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:**

**(i) p(x) = 2x**

^{3}+ x^{2}– 2x – 1, g(x) = x + 1

**(ii) p(x) = x**

^{3}+ 3x^{2}+ 3x + 1, g(x) = x + 2

**(iii) p(x) = x**

^{3}– 4x^{2}+ x + 6, g(x) = x – 3

Answer:

(i) p(x) = 2x

^{3}+ x^{2}– 2x – 1, g(x) = x + 1
Zero of x+1 is -1. If g(x) is a factor of p(x) then p(-1) = 0

⇒ p(-1) = 2(-1)

^{3}+ (-1)^{2}– 2(-1) – 1 = -2 + 1 + 2 - 1 = 0
∴ g(x) is a factor of polynomial p(x).

(ii) p(x) = x

^{3}+ 3x^{2}+ 3x + 1, g(x) = x + 2
Zero of x + 2 is -2. If g(x) is a factor of p(x) then p(-2) = 0

⇒ p(-2) = (-2)

^{3}+ 3(-2)^{2}+ 3(-2) + 1 = -8 + 12 - 6 + 1 = -1 ≠ 0
∴ g(x) is not a factor of polynomial p(x).

(iii) p(x) = x

^{3}– 4x^{2}+ x + 6, g(x) = x – 3
Root of x-2 is 3. If g(x) is a factor of p(x) then p(3) = 0

⇒ p(3) = (3)

^{3}– 4(3)^{2}+ 3 + 6 = 27 - 4(9) + 3 + 6 = 27 - 36 + 3 + 6 = 0
∴ g(x) is a factor of polynomial p(x).

**Q3: Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:**

**(i) p(x) = x**

^{2}+ x + k

**(ii) p(x) = 2x**

^{2}+ kx + √2

**(iii) p(x) = kx**

^{2}– √2x + 1

**(iv) p(x) = kx**

^{2}– 3x + k
Answer:

According to factor theorem, if x -1 is a factor of p(x), then p(1) = 0.

(i) p(x) = x

^{2}+ x + k
⇒ p(1) = (1)

^{2}+ 1 + k = 0
⇒ 1 + 1 + k = 0

⇒

**k = -2**
(ii) p(x) = 2x

^{2}+ kx + √2
⇒ p(1) = 0

⇒ p(1) = 2(1)

^{2}+ k(1) + √2 = 0
⇒ 2 + k + √2 = 0

⇒ k = -2 - √2 =

**-(2 +√2)**
(iii) p(x) = kx

^{2}– √2x + 1
⇒ p(1) = 0

⇒ p(1) = k(1)

^{2}– √2(1) + 1 = 0
⇒ k - √2 + 1

⇒ k =

**√2 - 1**
(iv) p(x) = kx

^{2}– 3x + k
⇒ p(1) = 0

⇒ p(1) = k(1)

^{2}– 3(1) + k = 0
⇒ k - 3 + k = 0

⇒ 2k = 3

⇒

**k = 3/2****Q4. Factorise:**

**(i) 12x**

^{2}– 7x + 1**(ii) 2x**

^{2}+ 7x + 3**(iii) 6x**

^{2}+ 5x – 6**(iv) 3x**

^{2}– x – 4
Answer:

(i) 12x

^{2}– 7x + 1
Method I: By splitting method, let us find out two number p and q such that pq = 12
× 1

and p + q = -7

i.e. p = -4 and q = -3

⇒ = 12x

^{2}– 4x -3x + 1
= 4x(3x - 1) - 1(3x -1)

= (4x -1)(3x -1) ...(answer)

Method II: By factor theorem.

(ii) 2x

^{2}+ 7x + 3

Let us find out two number p and q such that pq = 2
× 3 = 6 and p + q = 7

i.e. p = 6 and q = 1

⇒ = 2x

^{2}+ 6x + x + 3
= 2x (x + 3) + 1(x + 3)

= (2x + 1)(x + 3) ... answer
(iii) 6x

^{2}+ 5x – 6
Let us find out two number p and q such that pq = 6
× -6 = -36 and p + q = 5

i.e. p = 9 and q = -4

∴ 6x

^{2}+ 5x – 6 = 6x^{2}+ 9x -4x – 6
= 3x(2x + 3) - 2(2x + 3)

= (2x + 3)(3x - 2) ... answer

(iv) 3x

^{2}– x – 4

Let us find out two number p and q such that pq = 3
× -4 = -12 and p + q = -1

i.e. p = 3 and q = -4

∴ 3x

^{2}– x – 4 = 3x^{2}+ 3x –4x – 4
= 3x(x + 1) -4(x+1)

= (x+1)(3x - 4) ... answer

**Q5. Factorise:**

**(i) x**

^{3}– 2x^{2}– x + 2**(ii) x**

^{3}– 3x^{2}– 9x – 5**(iii) x**

^{3}+ 13x^{2}+ 32x + 20**(iv) 2y**

^{3}+ y^{2}– 2y – 1
Answer:

(i) Let p(x) = x

^{3}– 2x^{2}– x + 2
Here constant term is 2. Possible factors of 2 are: ±1, ±2

By trial method, p(2) = (2)

^{3}– 2(2)^{2}– 2 + 2 = 8 - 8 - 2 + 2 = 0
∴ (x -2 ) is factor of p(x).

P(x) ÷ (x -2) =

x^{2}- 1

▁▁▁▁▁▁▁▁▁▁▁▁▁ x -2 ) x^{3}– 2x^{2}– x + 2 x^{3}- 2x^{3}

- + ▔▔▔▔▔▔▔▔ -x + 2 -x + 2 + - ▔▔▔▔▔ 0

Since, Dividend = Divisor × Quotient + Remainder

∴ x

^{3}– 2x^{2}– x + 2 = (x-2)(x^{2}– 1) + 0
= (x - 2)( x

^{2}-x + x– 1)
= (x -2) [x(x-1) + 1 (x-1)]

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

(ii) Let f(x) = x

^{3}– 3x^{2}– 9x – 5
Here constant is 5, Possible factors of 5 are ±1 and ±5

By trial method, p(5) = (5)

^{3}– 3(5)^{2}– 9(5) – 5 = 125 - 75 - 45 - 5 = 0
∴ (x - 5) is factor of polynomial f(x).

Let us find out quotient = f(x) ÷ (x -5)

x^{2 }+ 2x + 1 ▁▁▁▁▁▁▁▁▁▁ x - 5) x^{3}– 3x^{2}– 9x – 5 x^{3}- 5x^{2}

- +

▔▔▔▔▔▔

0 + 2x^{2}- 9x - 5

2x^{2}- 10x

- +

▔▔▔▔▔▔▔▔

x - 5

x - 5

▔▔▔▔▔

0

Since, Dividend = Divisor × Quotient + Remainder

∴ x

^{3}– 3x^{2}– 9x – 5 = (x - 5)(x^{2 }+ 2x + 1)
Applying splitting method,

= (x -5)( x

^{2 }+ x + x + 1)
= (x -5)[x(x+1) +1(x+1)]

= (x-5)(x+1)(x+1)

= (x-5)(x+1)^{2}

**Method II**:

x

^{3}– 3x^{2}– 9x – 5 = x^{3}– 5x^{2}+ 2x^{2}– 9x – 5
= x

^{2}(x – 5)^{}+ 2x^{2}– 10x + x – 5
= x

^{2}(x – 5) + 2x(x - 5) + 1 (x -5)
= (x - 5)( x

^{2}+ 2x + 1)
= (x -5)[x(x+1) +1(x+1)]

= (x-5)(x+1)(x+1)

= (x-5)(x+1)^{2}

(iii) Let p(x) = x

^{3}+ 13x^{2}+ 32x + 20
Here constant term is 20. Its factors are: ±1, ±2, ±4, ±5 etc.

By trial method, p(-2) = (-2)

^{3}+ 13(-2)^{2}+ 32(-2) + 20
= -8 + 52 -64 + 20 = 0

∴ x + 2 is a factor of polynomial p(x).

⇒ x

^{3}+ 13x^{2}+ 32x + 20 = x^{3}+ 2x^{2}+ 11x^{2}+ 22x + 10x + 20
= x

^{2}(x + 2) + 11x(x + 2) + 10(x+2)
Taking (x + 2) common, we get

= (x + 2)( x

^{2}+ 11x + 10)
(Note: you can use long division method also).

= (x + 2)(x

^{2}+ x + 10x + 10)
= (x+2)[x(x+1) + 10(x+1)]

= (x+2)(x+10)(x+1)

(iv) Let p(y) = 2y

^{3}+ y^{2}– 2y – 1
Here the constant term is 1. Possible factors are: ±1

By trial method, p(1) = 2(1)

So (y-1) is a factor of p(y)

Quotient = p(y) ÷ (y-1)

By trial method, p(1) = 2(1)

^{3}+ (1)^{2}– 2(1) – 1 = 2 + 1 - 2 - 1 = 0So (y-1) is a factor of p(y)

Quotient = p(y) ÷ (y-1)

2y^{2 }+ 3y + 1

_____________________ y -1 ) 2y^{3}+ y^{2}– 2y – 1 2y^{3}-2y^{2}- + ▔▔▔▔▔▔▔▔▔▔▔▔

3y^{2}- 2y - 1

3y^{2}- 3y

^{ - +}

▔▔▔▔▔▔▔▔▔▔▔▔

y - 1

y - 1- +▔▔▔▔▔▔▔▔▔▔▔▔0

Since, Dividend = Divisor × Quotient + Remainder

2y

= (y-1)(2y

= (y-1)(2y(y+1) + 1(y + 1))

= (y-1)(y+1)(2y + 1)

Applying remainder theorem, if (x - 1) is factor of p(x) then p(1) = 0.

⇒ p(x) = ax

⇒ p(1) = a(1)

⇒ a + b + c + d = 0

We can say that,

Answer: Adding the co-efficients i.e. 5 + (-4) + (-2) + 1 = 6 - 6 = 0

⇒ (x - 1) is a factor of the given polynomial.

2y

^{3}+ y^{2}– 2y – 1 = (y -1)(2y^{2}+ 3y + 1)= (y-1)(2y

^{2}+ 2y + y + 1)= (y-1)(2y(y+1) + 1(y + 1))

= (y-1)(y+1)(2y + 1)

**Criterion to check if (x - 1) is a factor of polynomial p(x) = ax**

^{3}+ bx^{2}+ cx + d, where a,b,c,d ∈ R, a≠0Applying remainder theorem, if (x - 1) is factor of p(x) then p(1) = 0.

⇒ p(x) = ax

^{3}+ bx

^{2}+ cx + d

⇒ p(1) = a(1)

^{3}+ b(1)

^{2}+ c(1) + d = 0

⇒ a + b + c + d = 0

We can say that,

❝(x - 1) is a factor of polynomial p(x), if the sum of all the coefficients of polynomial p(x) is zero.❞

(✏ By by applying similar logic, can you find criterion to check if (x + 1) is a factor?)

**Q6:Check if the polynomial 5x**^{4}- 4x^{3}- 2x + 1 has (x - 1) its factor.Answer: Adding the co-efficients i.e. 5 + (-4) + (-2) + 1 = 6 - 6 = 0

⇒ (x - 1) is a factor of the given polynomial.

can u suggest some other ways for question 5? i hav done another method but cant do same for question 5, 2nd

ReplyDeleteSee the update blog page. Method II is added for 5(ii). Hope it helps!

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Deletei want to understand factorise questions.

ReplyDeleteits simple just do with the help of long division and after wise factorise it

ReplyDeletewow! thanks a lot

ReplyDeleteThanks a lot, You really make it easy for me and may be for everyone...............

ReplyDeleteI'm in eighth but i studied the topic outta curiousity . however I was stuck on a question . This thing really helped !

ReplyDeletewhere can i find the other method of Q5

ReplyDelete