Find two monomials whose product equals -20a^4b², and whose sum is a monomial with a coefficient of 1.

There are multiple ways to represent the product of two monomials as the product of two different monomials, but one possible way to find two monomials whose product equals -20a^4b² and whose sum is a monomial with a coefficient of 1 is:

-20a^4b² = (-1)(20a^4b²) = -(5a^4b²)(4b)

So, one possible pair of monomials is -5a^4b² and 4b. Their sum is -5a^4b² + 4b = -5a^4b² + 4b^1 = -5a^4b² + 4 with a coefficient of 1

Another possible pair of monomials that multiply to -20a^4b² and whose sum is a monomial with a coefficient of 1 is:

-20a^4b² = (-1)(20a^4b²) = -(10a^4b²)(2b)

So, another pair of monomials is -10a^4b² and 2b. Their sum is -10a^4b² + 2b = -10a^4b² + 2b^1 = -10a^4b² + 2 with a coefficient of 1

semsee45 semsee45 · Answer 2 · 2023-01-15T01:34:09+01:00

Answer:

-4a²b

5a²b

Step-by-step explanation:

A monomial is a polynomial that has one term only but can have multiple variables.

Given monomial:

[tex]-20a^4b^2[/tex]

The coefficient of the given monomial is -20.

Therefore, we need to find two numbers that multiply to -20 and sum to 1.

Factors of -20:

-1 and 20
-2 and 10
-4 and 5
-5 and 4
-10 and 2
-20 and 1

Therefore, the two numbers that multiply to -20 and sum to 1 are:

-4 and 5

Rewrite -20 as the product of -4 and 5:

[tex]\implies -4 \cdot 5 \cdot a^4b^2[/tex]

Rewrite the exponents as sums of equal numbers:

[tex]\implies -4 \cdot 5 \cdot a^{2+2} \cdot b^{1+1}[/tex]

[tex]\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c[/tex]

[tex]\implies -4 \cdot 5 \cdot a^2 \cdot a^2\cdot b^{1}\cdot b^{1}[/tex]

Rearrange as the product of two monomials with the same variables:

[tex]\implies -4a^2 b^{1}\cdot 5 a^2 b^{1}[/tex]

[tex]\implies -4a^2 b\cdot 5 a^2 b[/tex]

Therefore, the two monomials whose product equals -20a⁴b², and whose sum is a monomial with a coefficient of 1 are:

-4a²b
5a²b

Check the sum of the two found monomials:

[tex]\begin{aligned}\implies -4a^2b+5a^2b&=(-4+5)a^2b\\&=(1)a^2b\\&=a^2b\end{aligned}[/tex]

Thus proving that the sum of the monomials has a coefficient of 1.