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Yanka [14]
3 years ago
8

Replace ∗ with a monomial so that the result is an identity:

Mathematics
1 answer:
Mumz [18]3 years ago
7 0

QUESTION 1

Taking the right hand side we have;

4a^2-b^2

We can rewrite this as;

(2a)^2-b^2

We apply difference of two squares.

p^2-q^2=(p+q)(p-q)

(2a-b)(2a+b)

\therefore (2a-b)(2a+b)=4a^2-b^2

QUESTION 2

Taking the right hand side we have;

25x^2-0.16y^4

We can rewrite this as;

(5x)^2-(0.4y^2)^2

We apply difference of two squares.

p^2-q^2=(p+q)(p-q)

(5x-0.4y^2)(5x+0.4y^2)

\therefore (5x-0.4y^2)(5x+0.5y^2)=25x^2-0.16y^4

QUESTION 3

Taking the right hand side of the given equation, we have;

121a^{10}-b^8

We rewrite this as;

(11a^{5})^2-(b^4)^2

\therefore (11a^{5})^2-(b^4)^2=(11a^5-b^4)(11a^5+b^4)

QUESTION 4

From the RHS;

16y^2-9x^2

This implies that;

(4y)^2-(3x)^2

(4y)^2-(3x)^2=(4y-3x)(4y+3x)

QUESTION 5

From the left hand side, we have

100m^4-4n^6

This implies that;

(10m^2)^2-(2n^3)^2

Using difference of two squares, we have;

(10m^2)^2-(2n^3)^2=(10m^2-2n^3)(2n^3+10m^2)

QUESTION 6

From the LHS;

m^4-225m^{10}

We rewrite to obtain;

(m^2)^2-(15c^5)^2

Using difference of two squares, we obtain;

(m^2)^2-(15c^5)^2=(m^2-15c^5)(15c^5+m^2)

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