# 9th Class Math Chapter 6 Algebraic Manipulations Short Questions Answer

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## 9th Class Math Chapter 6 Algebraic Manipulations Short Questions Answer

Solution:

8*x*⁴ – 128 = 8(*x*⁴ – 16) = 8(*x*⁴ – 4²)

= 8(*x*² – 2²)(*x*² + 2²)

H.C.F. = 2³(*x* – 2)(*x* + 2)(*x*² + 4)

12*x*³ – 96 = 12(*x*³ – 8) = 2² x 3(*x*³ – 2³)

= 2² x 3(*x* – 2)(*x*² + 2*x*² + 4)

H.C.F. = 2²(*x* – 2)

= 4(*x* – 2)

Solution:

12*x*² – 75 = 3(4*x*² – 25) = 3[(2*x*)² – 5²]

= 3(2*x* – 5) (2*x* + 5)

6*x*² – 13*x* – 5 = 6*x*² – 15*x *+ 2*x* – 5

= 3*x *(2*x* – 5) (3*x* + 1)

= (2*x* – 5) (3*x* + 1)

4*x*² – 20*x* + 25 = (2*x*)² – 2(2*x*)(5) + 25

= 3*x *(2*x* – 5) (3*x* + 1)

= (2*x* – 5)²

So, L.C.M. = 3(2*x* + 5) (3*x* + 1) (2*x* – 5)²

Solution:

Let *p*(*x*) = *x*⁴ + 3*x*³ + 5*x*² + 26*x* + 56

*Q*(*x*) = *x*⁴ + 2*x*³ – 4*x*² – *x* + 28

H.C.F. = *x*² + 5*x* + 7

L.C.M. = *p*(*x*) x *Q*(*x*)/H.C.F.

= (*x*⁴ + 3*x*³ + 5*x*² + 26*x* + 56)(* x*⁴ + 2*x*³ – 4*x*² – *x* + 28)/*x*² + 5*x* + 7

Now dividing

So, L.C.M.

= (*x*² + 5*x* + 7)(*x*² + 2*x* + 8)(x⁴ + 2*x*³ – 4*x*62 – *x* + 28)/*x*²+5*x* +7

= (*x*² + 2*x* + 8)(x⁴ + 2*x*³ – 4*x*62 – *x* + 28)

Ans: Solution:

= *x*² + 1/*x*² + 2 + 10 (*x* + 1/*x*) + 27 – 2

= (*x* + 1/*x*)² + 10 (*x* + 1/*x*) + 25

= (*x* + 1/*x*)² + 2 (*x* + 1/*x*) + 5 + 5²

= [(*x* + 1/*x*) + 5]²

Required square root is ± [(*x* + 1/*x*) + 5]

Solution: