Question

Replace ∗ with a monomial so that the result is an identity:
1. (2a+*)(2a-*)=4a^2-b^2
2. (5x+*)(5x-*)=25x^2-0.16y^4
3. (*-b^4)(b^4+*)=121a^10-b^8
4. (*-3x)(*+3x)=16y^2-9x^2
5. 100m^4-4n^6=(10m^2-*)(*+10m^2)
6. m^4–225c^10 = (m^2 − *)(* +m^2)

Answer 1

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;

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