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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.5

Question 1.

Factorise the following expressions:

(i) 2a² + 4a²b + 8a²c

(ii) ab – ac – mb + mc

Solution:

(i) 2a² + 4a²b + 8a²c = 2a²(1 + 2b + 4c)

(ii) ab – ac – mb + mc = a(b – c) – m(b – c)

= (b – c) (a – m)

Question 2.

Factorise the following expressions:

(i) x² + 4x + 4

(ii) 3a² – 24ab + 48b²

(iii) x^{5} – 16x

(iv) m^{2} + \(\frac{1}{m^2}\) – 23

(v) 6 – 216x^{2}

(vi) a^{2} + \(\frac{1}{a^2}\) – 18

Solution:

(i) x^{2} + 4x + 4 = x^{2} + 2 × x × 2 + 2^{2} [a^{2} + 2ab+ b^{2} = (a + b)^{2}]

= (x + 2)^{2}

(ii) 3a^{2} – 24ab + 48b^{2} = 3[a^{2} – 8ab + 16b^{2}]

= 3[a^{2} – 2 × a × 4b + (4b)^{2}]

= 3(a- 4b)^{2} [a^{2} – 2ab + b^{2} = (a – b)^{2}]

(iii) x^{5} – 16x = x[x^{4} – 16] [a^{2} – b^{2} = (a + b) (a – b)]

= x[(x^{2})^{2} – 4^{2}]

= x(x^{2} + 4) (x^{2} – 4)

= x(x^{2} + 4) (x^{2} – 2^{2})

= x(x^{2}+ 4) (x + 2) (x – 2)

(iv) m^{2} + \(\frac{1}{m^2}\) – 23 = [Add + 2 and – 2 to make -23 as -25]

= m^{2} + \(\frac{1}{m^2}\) + 2 – 2 – 23 = m^{2} + \(\frac{1}{m^2}\) + 2 – 25

= m^{2} + \(\frac{1}{m^2}\) + 2 × m × \(\frac{1}{m}\) – 5^{2}

(v) 6 – 216x^{2} = 6[1 – 36x^{2}]

= 6[1 – (6x)^{2}] [a^{2} – b^{2} = (a + b)(a – b)]

= 6(1 + 6x) (1 – 6x)

(vi) a^{2} + \(\frac{1}{a^2}\) – 18 = a^{2} + \(\frac{1}{a^2}\) – 2 + 2 – 18

(add -2 and +2 to make 18 as 16)

= a^{2} + \(\frac{1}{a^2}\) – 2 × a × \(\frac{1}{a}\) – 16 [a^{2} + b^{2} – 2ab = (a-b)(a- b)^{2}]

Question 3.

Factorise the following expressions:

(i) 4x^{2} + 9y^{2} + 25z^{2} + 12xy + 30yz + 20xz

Solution:

[a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac = (a + b + c)^{2}]

= (2x)^{2} + (3y)^{2} + (5z)^{2} + 2(2x) (3y) + 2(3y) (5z) + 2(5z) (2x)

= (2x + 3y + 5z)^{2}

(ii) 25x^{2} + 4y^{2} + 9z^{2} – 20xy + 12yz – 30xz

Solution:

= (5x)^{2} + (2y)^{2} + (3z)^{2} + 2(5x) (-2y) + 2(-2y) (- 3z) + 2(-3z) (5x)

= (5x – 2y – 3z)^{2}

Question 4.

Factorise the following expressions:

(i) 8x^{3} + 125y^{3}

(ii) 27x^{3} – 8y^{3}

(iii) a^{6} – 64

Solution:

(i) 8x^{3} + 125y^{3} = (2x)^{3} + (5y)^{3} [a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})

= (2x + 5y) [(2x)^{2} – (2x) (5y) + (5y)^{2}]

= (2x + 5y) (4x^{2} – 10xy + 25y^{2})

(ii) 27x^{3} – 8y^{3} = (3x)^{3} – (2y)^{3} [a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2})]

= (3x – 2y) [(3x)^{2} + (3x) (2y) + (2y)^{2}]

= (3x – 2y) (9x^{2} + 6xy + Ay^{2})

(iii) a^{6} – 64 = a^{6} – 2^{6}

= (a^{2})^{3} – (22)^{3} [a^{2} – b^{3} = (a- b) + (a^{2} + ab + b^{2})]

= (a^{2} – 22) [(a^{2})^{2} + (a^{2}) (2^{2}) + (2^{2})^{2}]

= (a + 2) (a – 2) (a^{4} + 4a^{2} + 16)

= (a + 2) (a – 2) [(a^{2})^{2} + 4^{2} + 8a^{2 }– 4a^{2}]

= (a + 2)(a- 2) [(a^{2} + 4)^{2 }– (2a)^{2}] {a^{2} – b^{2} = (a + b) (a – b)}

= (a + 2) (a – 2) [(a^{2} + 4 + 2a) (a^{2} + 4 – 2a)

= (a + 2) (a – 2) (a^{2} + 2a + 4) (a^{2} – 2a + 4)

Question 5.

Factorize the following

(i) x^{3} + 8y^{3} + 6xy – 1

(ii) l^{3} – 8m^{3} – 27n^{3} – 18lmn

Using the formula [a^{3} + b^{3} + c^{3} – 3abc] = (a + b + c) (a^{2} + b^{2} + c^{2} – ab – bc – ac)

Solution:

(i) x^{3} + 8y^{2} + 6xy – 1 = -(-x^{3} – 8y^{3} – 6xy + 1)

= – (-x^{3} – 8y^{3} + 1 – 6xy)

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

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

= (x + 2y – 1)(x^{2} + 4y^{2} + 1 – 2xy + 2y + x)

(ii) l^{3} – 8m^{3} – 27n^{3} – 18lmn

= l^{3} + (-2m)^{3} + (-3n)^{2} – 3(l) (-2m) (-3n)

= (l – 2m – 3n) [l^{2} + (-2m)^{2} + (-3n)^{2} -1 (-2m)] – (-2m)(-3n) – (-3n)(l)

= (l – 2m – 3n) (l^{2} + 4m^{2} + 9n^{2} + 2lm – 6mn + 3ln)