How do you factor $9 t^{2} + 16 t + 4$ $9t ^ { 2} + 16t + 4$ ?

Question

How do you factor $9 t^{2} + 16 t + 4$ $9t ^ { 2} + 16t + 4$ ?

1 Answer

Impact of this question

1141 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-14T19:16:58

$\frac{1}{9} (9 t + 8 - 2 \sqrt{7}) (9 t + 8 + 2 \sqrt{7})$ $1/9(9t+8-2sqrt7)(9t+8+2sqrt7)$

Explanation:

$9 t^{2} + 16 t + 4 = 3^{2} t^{2} + 2 \cdot 3 \cdot (\frac{8}{3}) t + {(\frac{8}{3})}^{2} - {(\frac{8}{3})}^{2} + 4$ $9t^2+16t+4 = 3^2t^2 + 2*3*(8/3)t+(8/3)^2-(8/3)^2+4$
$= (3^{2} t^{2} + 2 \cdot 3 \cdot (\frac{8}{3}) t + {(\frac{8}{3})}^{2}) - {(\frac{8}{3})}^{2} + 4$ $= (3^2t^2 + 2*3*(8/3)t+(8/3)^2)-(8/3)^2+4$
$= {(3 t + \frac{8}{3})}^{2} - 4 {(\frac{4}{3})}^{2} + 4$ $= (3t+8/3)^2-4(4/3)^2+4$
$= {(3 t + \frac{8}{3})}^{2} + 4 (1 - \frac{16}{9})$ $=(3t+8/3)^2+4(1-16/9)$
$= {(3 t + \frac{8}{3})}^{2} + 4 (\frac{9 - 16}{9})$ $=(3t+8/3)^2+4((9-16)/9)$
$= {(3 t + \frac{8}{3})}^{2} - \frac{4}{9} \cdot 7$ $=(3t+8/3)^2-4/9*7$
$= {(3 t + \frac{8}{3})}^{2} - {(\frac{2 \sqrt{7}}{3})}^{2}$ $=(3t+8/3)^2-((2sqrt7)/3)^2$
$= (3 t + \frac{8}{3} - \frac{2 \sqrt{7}}{3}) (3 t + \frac{8}{3} + \frac{2 \sqrt{7}}{3})$ $=(3t+8/3-(2sqrt7)/3)(3t+8/3+(2sqrt7)/3)$

Science

Math

Humanities

... and beyond

How do you factor $9 t^{2} + 16 t + 4$ $9t ^ { 2} + 16t + 4$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor 9t2+16t+49t ^ { 2} + 16t + 4?

1 Answer

Explanation:

Related questions

Impact of this question

How do you factor $9 t^{2} + 16 t + 4$ $9t ^ { 2} + 16t + 4$ ?