Factorize- 75-a-b-2 - 48-a-b-2-3-a - 9b-9a - b-3-a - 3b-3a - b-3-a - 9b-9a - b-3-a - 9b-9a - b-

The correct option is D 3(a + 9b)(9a + b)) 75(a+b)^2 48(a b)^2= nbsp;3{25(a+b)^2 16(a b)^2} = nbsp;3{[5(a+b)]^2 [4(a b)]^2} Using the identity:a^2 b ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Factorisation of Algebraic Identities

Factorize: 75...

Question

Factorize: $75 (a + b)^{2} - 48 (a - b)^{2}$ .

$3 (a + 9 b) (9 a + b))$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$3 (a - 9 b) (9 a - b))$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$3 (a + 3 b) (3 a + b))$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$3 (a - 9 b) (9 a + b))$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$3 (a + 9 b) (9 a + b))$

$75 (a + b)^{2} - 48 (a - b)^{2} = 3 {25 (a + b)^{2} - 16 (a - b)^{2}}$
$= 3 {[5 (a + b)]^{2} - [4 (a - b)]^{2}}$

Using the identity: $a^{2} - b^{2} = (a - b) (a + b)$
$= 3 {[5 (a + b) - 4 (a - b)] [5 (a + b) + 4 (a - b)]}$

$= 3 {(5 a + 5 b - 4 a + 4 b) (5 a + 5 b + 4 a - 4 b)}$

$\Rightarrow 3 (a + 9 b) (9 a + b)$

Suggest Corrections

Similar questions

$(7 a^{2} - 63 b^{2}) = ? (a) (7 a - 9 b) (9 a + 7 b) (b) (7 a - 9 b) (7 a + 9 b) (c) 9 (a - 3 b) (a + 3 b) (d) 7 (a - 3 b) (a + 3 b)$

Q. Tick (✓) the correct answer:
(7a² − 63b²) = ?
(a) (7a − 9b)(9a + 7b)
(b) (7a − 9b)(7a + 9b)
(c) 9(a − 3b)(a + 3b)
(d) 7(a − 3b)(a + 3b)

Q. Without expanding, show that the values of each of the following determinants are zero:
(i)

|\begin{matrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{matrix}|

(ii)

|\begin{matrix} 6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2 \end{matrix}|

(iii)

|\begin{matrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{matrix}|

(iv)

|\begin{matrix} 1 / a & a^{2} & b c \\ 1 / b & b^{2} & a c \\ 1 / c & c^{2} & a b \end{matrix}|

(v)

|\begin{matrix} a + b & 2 a + b & 3 a + b \\ 2 a + b & 3 a + b & 4 a + b \\ 4 a + b & 5 a + b & 6 a + b \end{matrix}|

(vi)

|\begin{matrix} 1 & a & a^{2} - b c \\ 1 & b & b^{2} - a c \\ 1 & c & c^{2} - a b \end{matrix}|

(vii)

|\begin{matrix} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{matrix}|

(viii)

|\begin{matrix} 0 & x & y \\ - x & 0 & z \\ - y & - z & 0 \end{matrix}|

(ix)

|\begin{matrix} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{matrix}|

(x)

|\begin{matrix} 1^{2} & 2^{2} & 3^{2} & 4^{2} \\ 2^{2} & 3^{2} & 4^{2} & 5^{2} \\ 3^{2} & 4^{2} & 5^{2} & 6^{2} \\ 4^{2} & 5^{2} & 6^{2} & 7^{2} \end{matrix}|

(xi)

|\begin{matrix} a & b & c \\ a + 2 x & b + 2 y & c + 2 z \\ x & y & z \end{matrix}|

Q. Without expanding, show that the values of each of the following determinants are zero:
(i)

|\begin{matrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{matrix}|

(ii)

|\begin{matrix} 6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2 \end{matrix}|

(iii)

|\begin{matrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{matrix}|

(iv)

|\begin{matrix} 1 / a & a^{2} & b c \\ 1 / b & b^{2} & a c \\ 1 / c & c^{2} & a b \end{matrix}|

(v)

|\begin{matrix} a + b & 2 a + b & 3 a + b \\ 2 a + b & 3 a + b & 4 a + b \\ 4 a + b & 5 a + b & 6 a + b \end{matrix}|

(vi)

|\begin{matrix} 1 & a & a^{2} - b c \\ 1 & b & b^{2} - a c \\ 1 & c & c^{2} - a b \end{matrix}|

(vii)

|\begin{matrix} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{matrix}|

(viii)

|\begin{matrix} 0 & x & y \\ - x & 0 & z \\ - y & - z & 0 \end{matrix}|

(ix)

|\begin{matrix} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{matrix}|

(x)

|\begin{matrix} 1^{2} & 2^{2} & 3^{2} & 4^{2} \\ 2^{2} & 3^{2} & 4^{2} & 5^{2} \\ 3^{2} & 4^{2} & 5^{2} & 6^{2} \\ 4^{2} & 5^{2} & 6^{2} & 7^{2} \end{matrix}|

(xi)

|\begin{matrix} a & b & c \\ a + 2 x & b + 2 y & c + 2 z \\ x & y & z \end{matrix}|

(xii)

|\begin{matrix} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix}|

(xiii)

|\begin{matrix} \sin α & \cos α & \cos (α + δ) \\ \sin β & \cos β & \cos (β + δ) \\ \sin γ & \cos γ & \cos (γ + δ) \end{matrix}|

(xiv)

|\begin{matrix} \sin^{2} 23 ° & \sin^{2} 67 ° & \cos 180 ° \\ - \sin^{2} 67 ° & - \sin^{2} 23 ° & \cos^{2} 180 ° \\ \cos 180 ° & \sin^{2} 23 ° & \sin^{2} 67 ° \end{matrix}|

(xv)

|\begin{matrix} \cos (x + y) & - \sin (x + y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y \end{matrix}|

(xvi)

|\begin{matrix} \sqrt{23} + \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{46} & 5 & \sqrt{10} \\ 3 + \sqrt{115} & \sqrt{15} & 5 \end{matrix}|

(xvii)

|\begin{matrix} \sin^{2} A & \cot A & 1 \\ \sin^{2} B & \cot B & 1 \\ \sin^{2} C & \cot C & 1 \end{matrix}|, where A, B, C are the angles of ∆ A B C .

Factorize: $75 (a + b)^{2} - 48 (a - b)^{2}$ .

Join BYJU'S Learning Program

Factorize: 75(a+b)2−48(a−b)2.

The correct option is A 3(a+9b)(9a+b)) 75(a+b)2−48(a−b)2=3{25(a+b)2−16(a−b)2} =3{[5(a+b)]2−[4(a−b)]2} Using the identity:a2−b2=(a−b)(a+b) =3{[5(a+b)−4(a−b)][5(a+b)+4(a−b)]} =3{(5a+5b−4a+4b)(5a+5b+4a−4b)} ⇒3(a+9b)(9a+b)

Factorize: $75 (a + b)^{2} - 48 (a - b)^{2}$ .