Question

simplify- 4c^2-36/8c^2-24c/12c+36/2c^2-6c
pls help :)

Answer 1

$\displaystyle \dfrac{\dfrac{4c^2-36}{8c^2-24c}}{\dfrac{12c+36}{2c^2 -6c}}\\\\\\=\left(\dfrac{4c^2 -36}{8c^2 -24c}\right) \cdot \left(\dfrac{2c^2 -6c}{12c+36}\right)\\\\\\=\left[\dfrac{4(c^2 -9)}{8(c^2-3)}\right]\cdot\left[\dfrac{2(c^2-3)}{12(c+3)}\right]\\\\\\=\dfrac{8(c^2-3^2)}{8 \cdot 12(c+3)}\\\\\\=\dfrac{(c+3)(c-3)}{12(c+3)}\\\\\\=\dfrac{c-3}{12}$

$\text{Hence the answer is d.}$

Answer 2

Answer:

d

Step-by-step explanation:

separate the 2 divisions and simplify each , that is

$\frac{4c^2-36}{8c^2-24c}$ ← factor numerator/ denominator

= $\frac{4(c^2-9)}{8c(c-3)}$ ← factor difference of squares on numerator

= $\frac{4(c-3)(c+3)}{8c(c-3)}$ ← cancel (c - 3) and 4/8 on numerator/ denominator

= $\frac{c+3}{2c}$

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$\frac{12c+36}{2c^2-6c}$ ← factor numerator/ denominator

= $\frac{12(c+3)}{2c(c-3)}$ ← cancel 12 and 2 on numerator/ denominator

= $\frac{6(c+3)}{c(c-3)}$

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to divide the top fraction by the lower fraction

leave top fraction, change division to multiplication, turn lower fraction upside down.

$\frac{c+3}{2c}$ × $\frac{c(c-3)}{6(c+3)}$ ← cancel (c + 3) and c on numerator/ denominator

= $\frac{c-3}{2(6)}$

= $\frac{c-3}{12}$ → d