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nordsb [41]
3 years ago
10

Find the greatest common factor of 12a^3b, 16a^2b^2, and 36ab^3

Mathematics
1 answer:
scoundrel [369]3 years ago
3 0

<u>Answer: </u>

The greatest common factor of 12 a^{3} b, 16 a^{2} b^{2}, 36 a b^{3} \text { is } 4 a b

<u>Solution: </u>

To find the greatest common factor we have to find the prime factors of individual numbers and then find the number which is common to each given number.

Here the numbers are 12 a^{3} b, 16 a^{2} b^{2}, 36 a b^{3}

Let us find out the prime factors of each number .

\begin{array}{l}{\text { Prime factors of } 12 a^{3} b=2 \times 2 \times 3 \times a \times a \times a \times b} \\ {\text { Prime factors of } 16 a^{2} b^{2}=2 \times 2 \times 2 \times 2 \times a \times a \times a \times b \times b} \\ {\text { Prime factors of } 36 a b^{3}=2 \times 2 \times 3 \times 3 \times a \times b \times b \times b}\end{array}

We can see that 2 \times 2 \times \mathrm{a} \times \mathrm{b}  is common to all the given numbers 12 a^{3} b, 16 a^{2} b^{2}, 36 a b^{3}

Therefore the greatest common factor of 12 a^{3} b, 16 a^{2} b^{2}, 36 a b^{3} \text { is } 2 \times 2 \times a \times b=4 a b

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Which of the binomials below is a factor of this trinomial? x^2 + 3x - 4A. x+4B. x+8C. x-4D. x^2+4
Usimov [2.4K]

Given:

The equation is:

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\begin{gathered} =x^2+3x-4 \\  \\ =x^2+4x-x-4 \\  \end{gathered}

Then the factor is:

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