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Exercise 2.3 Class 9 Maths | Factor Theorem
Exercise 2.3 Class 9 Maths | Factor Theorem by ‪@rajansir07‬

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Class 9 Chapter 2 – Polynomials Exercise 2.4
1. Determine which of the following polynomials has (x + 1) a factor:

(i) x3+x2+x+1

Solution:

Let p(x) = x3+x2+x+1

The zero of x+1 is -1. [x+1 = 0 means x = -1]

p(−1) = (−1)3+(−1)2+(−1)+1

= −1+1−1+1

= 0

∴By factor theorem, x+1 is a factor of x3+x2+x+1

(ii) x4+x3+x2+x+1

Solution:

Let p(x)= x4+x3+x2+x+1

The zero of x+1 is -1. . [x+1= 0 means x = -1]

p(−1) = (−1)4+(−1)3+(−1)2+(−1)+1

= 1−1+1−1+1

= 1 ≠ 0

∴By factor theorem, x+1 is not a factor of x4 + x3 + x2 + x + 1

(iii) x4+3x3+3x2+x+1

Solution:

Let p(x)= x4+3x3+3x2+x+1

The zero of x+1 is -1.

p(−1)=(−1)4+3(−1)3+3(−1)2+(−1)+1

=1−3+3−1+1

=1 ≠ 0

∴By factor theorem, x+1 is not a factor of x4+3x3+3x2+x+1

(iv) x3 – x2– (2+√2)x +√2

Solution:

Let p(x) = x3–x2–(2+√2)x +√2

The zero of x+1 is -1.

p(−1) = (-1)3–(-1)2–(2+√2)(-1) + √2 = −1−1+2+√2+√2

= 2√2 ≠ 0

∴By factor theorem, x+1 is not a factor of x3–x2–(2+√2)x +√2
2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p(x) = 2x3+x2–2x–1, g(x) = x+1

Solution:

p(x) = 2x3+x2–2x–1, g(x) = x+1

g(x) = 0

⇒ x+1 = 0

⇒ x = −1

∴Zero of g(x) is -1.

Now,

p(−1) = 2(−1)3+(−1)2–2(−1)–1

= −2+1+2−1

= 0

∴By factor theorem, g(x) is a factor of p(x).

(ii) p(x)=x3+3x2+3x+1, g(x) = x+2

Solution:

p(x) = x3+3x2+3x+1, g(x) = x+2

g(x) = 0

⇒ x+2 = 0

⇒ x = −2

∴ Zero of g(x) is -2.

Now,

p(−2) = (−2)3+3(−2)2+3(−2)+1

= −8+12−6+1

= −1 ≠ 0

∴By factor theorem, g(x) is not a factor of p(x).

(iii) p(x)=x3–4x2+x+6, g(x) = x–3

Solution:

p(x) = x3–4x2+x+6, g(x) = x -3

g(x) = 0

⇒ x−3 = 0

⇒ x = 3

∴ Zero of g(x) is 3.

Now,

p(3) = (3)3−4(3)2+(3)+6

= 27−36+3+6

= 0

∴By factor theorem, g(x) is a factor of p(x
. Find the value of k, if x–1 is a factor of p(x) in each of the following cases:

(i) p(x) = x2+x+k

Solution:

If x-1 is a factor of p(x), then p(1) = 0

By Factor Theorem

⇒ (1)2+(1)+k = 0

⇒ 1+1+k = 0

⇒ 2+k = 0

⇒ k = −2

(ii) p(x) = 2x2+kx+√2

Solution:

If x-1 is a factor of p(x), then p(1)=0

⇒ 2(1)2+k(1)+√2 = 0

⇒ 2+k+√2 = 0

⇒ k = −(2+√2)

(iii) p(x) = kx2–√2x+1

Solution:

If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

⇒ k(1)2-√2(1)+1=0

⇒ k = √2-1

(iv) p(x)=kx2–3x+k

Solution:

If x-1 is a factor of p(x), then p(1) = 0

By Factor Theorem

⇒ k(1)2–3(1)+k = 0

⇒ k−3+k = 0

⇒ 2k−3 = 0

⇒ k= 3/2
Solution:

Let p(x) = x3–2x2–x+2

Factors of 2 are ±1 and ± 2

Now,

p(x) = x3–2x2–x+2

p(−1) = (−1)3–2(−1)2–(−1)+2

= −1−2+1+2

= 0
Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x2–3x+2) = (x+1)(x2–x–2x+2)

= (x+1)(x(x−1)−2(x−1))

= (x+1)(x−1)(x-2)

(ii) x3–3x2–9x–5

Solution:

Let p(x) = x3–3x2–9x–5

Factors of 5 are ±1 and ±5

By trial method, we find that

p(5) = 0

So, (x-5) is factor of p(x)

Now,

p(x) = x3–3x2–9x–5

p(5) = (5)3–3(5)2–9(5)–5

= 125−75−45−5

= 0
Now, Dividend = Divisor × Quotient + Remainder

(x−5)(x2+2x+1) = (x−5)(x2+x+x+1)

= (x−5)(x(x+1)+1(x+1))

= (x−5)(x+1)(x+1)

(iii) x3+13x2+32x+20

Solution:

Let p(x) = x3+13x2+32x+20

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By trial method, we find that

p(-1) = 0

So, (x+1) is factor of p(x)

Now,

p(x)= x3+13x2+32x+20

p(-1) = (−1)3+13(−1)2+32(−1)+20

= −1+13−32+20

= 0