Question

can anybody solve this sum. 80x^3 - 45x ​

Answer 1

Answer:

5x(4x+3)(4x-3)

Step-by-step explanation:

To solve this problem, first you have to use the distributive property of a(b+c)=ab+ac.

First, factor out by common term.

5x(16x²-9)

Then, solve.

16x²-9=(4x+3)(4x-3)

In conclusion, the correct answer is 5x(4x+3)(4x-3).

Answer 2

Answer:

$80x^3-45x:\quad 5x\left(4x+3\right)\left(4x-3\right)$

Step-by-step explanation:

$80x^3-45x$

$\black{\mathrm{Factor\:out\:common\:term\:}5x:}$

$80x^3-45x$

$\gray{\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c}$

$\gray{x^3=x^2x}$

$=80x^2x-45x$

$\gray{\mathrm{Rewrite\:}45\mathrm{\:as\:}5\cdot \:9}$

$\gray{\mathrm{Rewrite\:}80\mathrm{\:as\:}5\cdot \:16}$

$=5\cdot \:16x^2x-5\cdot \:9x$

$\gray{\mathrm{Factor\:out\:common\:term\:}5x}$

$=5x\left(16x^2-9\right)$

$\black{\mathrm{Factor}\:16x^2-9:}$

$16x^2-9$

$\gray{\mathrm{Rewrite\:}16x^2-9\mathrm{\:as\:}\left(4x\right)^2-3^2}$

$=\left(4x\right)^2-3^2$

$\gray{\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)}$

$\gray{\left(4x\right)^2-3^2=\left(4x+3\right)\left(4x-3\right)}$

$=\left(4x+3\right)\left(4x-3\right)$

$=5x\left(4x+3\right)\left(4x-3\right)$