3 years ago

What is the product? (6 r minus 1)(negative 8 r minus 3)

Mathematics

2 answers:

just olya [345]3 years ago

3 0

Answer:

The required product is $-48r^2-10r+3$

Step-by-step explanation:

Consider the provided expression.

$(6r-1)(-8r-3)$

Open the parentheses.

$6r\left(-8r\right)+6r\left(-3\right)+\left(-1\right)\left(-8r\right)+\left(-1\right)\left(-3\right)$

$-6\cdot \:8rr-6\cdot \:3r+1\cdot \:8r+1\cdot \:3$

$-48r^2-18r+8r+3$

Add the like terms.

$-48r^2-10r+3$

Hence, the required product is $-48r^2-10r+3$

Marta_Voda [28]3 years ago

3 0

Answer:

-47r²-10r+3

Step-by-step explanation:

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