KnowledgeBoat Logo
|

Mathematics

Factorise :

(i) x3 - 2x2 - x + 2

(ii) x3 - 3x2 - 9x - 5

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

(iv) 2y3 + y2 - 2y - 1

Polynomials

15 Likes

Answer

(i) x3 - 2x2 - x + 2

Let (x - a) is a factor, a = 1, -1, 2, -2…….

Putting a = 1

x - 1 = 0

x = 1

p(x) = x3 - 2x2 - x + 2

p(1) = (1)3 - 2 x (1)2 - 1 + 2

= 1 - 2 - 1 + 2

= 0

Hence, (x - 1) is the factor of x3 - 2x2 - x + 2

On dividing, x3 - 2x2 - x + 2 by (x - 1)

x−1)x2−x−2x−1)x3−2x2−x+2‾x−1)2+−x3−+x2‾x−1x3−2−x2−x+2x−1)x3−2−+x2+−x‾x−1)x3−2x2(3)−2x+2x−1)x3−2x2(31)−+2x+−2‾x−1)x3−2x2(31)−2x×\begin{array}{l} \phantom{x - 1)}{\quad x^2 -x - 2} \ x - 1\overline{\smash{\big)}\quad x^3 - 2x^2 - x + 2} \ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{+}{-}x^2} \ \phantom{{x - 1}x^3-2}-x^2 - x + 2 \ \phantom{{x - 1)}x^3-2}\underline{\underset{+}{-}x^2 \underset{-}{+} x} \ \phantom{{x - 1)}{x^3-2x^{2}(3)}}-2x + 2 \ \phantom{{x - 1)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}2x \underset{-}{+} 2} \ \phantom{{x - 1)}{x^3-2x^{2}(31)}{-2x}}\times \end{array}

we get quotient = x2 - x - 2

Factorising x2 - x - 2,

= x2 - x - 2

= x2 -(2 - 1)x - 2

= x2 - 2x + x -2

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

= (x + 1)(x - 2)

∴ x2 - x - 2 = (x + 1)(x - 2)

∴ p(x) = x3 - 2x2 - x + 2 = (x - 1) (x + 1) (x - 2)

So, factors are (x - 1) (x + 1) (x - 2)

(ii) x3 - 3x2 - 9x - 5

Let (x - a) is a factor, a = 1, -1, 2, -2…….

Putting a = -1

x + 1 = 0

x = -1

p(x) = x3 - 3x2 - 9x - 5

p(-1) = (-1)3 - 3 x (-1)2 - 9 x (-1) - 5

= -1 - 3 + 9 - 5

= -9 + 9

= 0

Hence, (x + 1) is a factor of x3 - 3x2 - 9x - 5

On dividing, x3 - 3x2 - 9x - 5 by (x + 1)

x+1)x2−4x−5x+1)x3−3x2−9x−5‾x−1)2+−x3+−x2‾x+1x3−2−4x2−9xx+1)x3−2−+4x2−+4x‾x+1)x3−2x2(3)−5x−5x+1)x3−2x2(31)−+5x−+5‾x+1)x3−2x2(31)−2×\begin{array}{l} \phantom{x + 1)}{\quad x^2 -4x - 5} \ x + 1\overline{\smash{\big)}\quad x^3 - 3x^2 - 9x - 5} \ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{-}{+}x^2} \ \phantom{{x + 1}x^3-2}-4x^2 - 9x \ \phantom{{x + 1)}x^3-2}\underline{\underset{+}{-}4x^2 \underset{+}{-} 4x} \ \phantom{{x + 1)}{x^3-2x^{2}(3)}}-5x - 5 \ \phantom{{x + 1)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}5x \underset{+}{-} 5} \ \phantom{{x + 1)}{x^3-2x^{2}(31)}{-2}}\times \end{array}

We get quotient = x2 - 4x - 5

Factorising x2 - 4x - 5

x2 -(5 - 1)x - 5

x2 -5x + x - 5

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

(x + 1) (x - 5)

∴ x2 - 4x - 5 = (x + 1) (x - 5)

∴ p(x) = x3 - 3x2 -9x - 5 = (x + 1) (x + 1) (x - 5)

So, factors are (x + 1) (x + 1) (x - 5)

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

Let (x - a) is a factor, a = 1, -1, 2, -2…….

Putting a = -1

x + 1 = 0

x = -1

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

p(-1) = (-1)3 + 13 x (-1)2 + 32 x (-1) + 20

= -1 + 13 - 32 + 20

= -33 + 33

= 0

Hence, (x + 1) is a factor of x3 + 13x2 + 32x + 20

On dividing, x3 + 13x2 + 32x + 20 by (x + 1)

x+1)x2+12x+20x+1)x3+13x2+32x+20‾x−1)2+−x3+−x2‾x+1x3−212x2+32x+20x+1)x31+−12x2+−12x‾x+1)x3−2x2(3)20x+20x+1)x3−2x23+−20x+−20‾x+1)x3−2x2(3)−2×\begin{array}{l} \phantom{x + 1)}{\quad x^2 +12x + 20} \ x + 1\overline{\smash{\big)}\quad x^3 + 13x^2 + 32x + 20} \ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{-}{+}x^2} \ \phantom{{x + 1}x^3-2}12x^2 + 32x + 20 \ \phantom{{x + 1)}x^31}\underline{\underset{-}{+}12x^2 \underset{-}{+} 12x} \ \phantom{{x + 1)}{x^3-2x^{2}(3)}}20x + 20 \ \phantom{{x + 1)}{x^3-2x^{2}3}}\underline{\underset{-}{+}20x \underset{-}{+} 20} \ \phantom{{x + 1)}{x^3-2x^{2}(3)}{-2}}\times \end{array}

We get quotient = x2 + 12x + 20

Factorising x2 + 12x + 20

x2 + 10x + 2x + 20

x(x + 10) + 2(x + 10)

(x + 10)(x + 2)

∴ x2 + 12x + 20 = (x + 10)(x + 2)

∴ p(x) = x3 + 13x2 + 32x + 20 = (x + 1) (x + 10) (x + 2)

So, factors are (x + 1) (x + 10) (x + 2)

(iv) 2y3 + y2 - 2y - 1

Let (y - a) is a factor, a = 1, -1, 2, -2…….

Putting a = 1

y - 1 = 0

y = 1

p(y) = 2y3 + y2 - 2y - 1

p(1) = 2 x (1)3 + (1)2 - 2 x 1 - 1

= 2 + 1 - 2 - 1

= 0

Hence, (y - 1) is a factor of 2y3 + y2 - 2y - 1

On dividing, 2y3 + y2 - 2y - 1 by (y - 1)

y−1)2y2+3y+1y−1)2y3+y2−2y−1‾y−1)+−2y3−+2y2‾y−1x3−2333y2−2yy−1)x3−2+−3y2−+3y‾y−1)x3−2x2(3333)y−1y−1)x3−2x23333+−y−+1‾y−1)x3−2x2(333)−2×\begin{array}{l} \phantom{y - 1)}{\quad 2y^2 + 3y + 1} \ y - 1\overline{\smash{\big)}\quad 2y^3 + y^2 - 2y - 1} \ \phantom{y - 1)}\underline{\underset{-}{+}2y^3 \underset{+}{-}2y^2} \ \phantom{{y - 1}x^3-233}3y^2 - 2y \ \phantom{{y - 1)}x^3-2}\underline{\underset{-}{+}3y^2 \underset{+}{-} 3y} \ \phantom{{y - 1)}{x^3-2x^{2}(3333)}}y - 1 \ \phantom{{y - 1)}{x^3-2x^{2}3333}}\underline{\underset{-}{+}y \underset{+}{-} 1} \ \phantom{{y - 1)}{x^3-2x^{2}(333)}{-2}}\times \end{array}

We get quotient = 2y2 + 3y + 1

Factorising 2y2 + 3y + 1

2y2 + 2y + y +1

2y(y + 1) + 1(y + 1)

(y + 1)(2y + 1)

∴ 2y2 + 3y + 1 = (y + 1)(2y + 1)

∴ p(y) = 2y3 + y2 - 2y - 1 = (y - 1)(y + 1)(2y + 1)

So, factors are (y - 1)(y + 1)(2y + 1)

Answered By

8 Likes

Related Questions