Question

−10t5 + 15t4 + 9t3 factor

Answer 1

Here \[t^3 \]is common.

Answer 2

Oh, we've done these quadratic forms before . . . ;-)

Answer 3

\[\LARGE{\color{red}{t^3(-10t^2+15t+9)}}\]
Then factor the eq in the brackets:)

Answer 4

\[−10t^5 + 15t^4 + 9t^3\]
\[t^3(−10t^2 + 15t + 9)\]
now just use the quadratic formula

Answer 5

\[(-b +/- \sqrt{(b^2 -4ac))})/2a\]

Answer 6

Quadratic formula isn't required for factoring. I think the quadratic part is actually prime.

Answer 7

\[\Huge{\color{blue}{\frac{-b- \sqrt{b^2-4ac}}{2a}}}\]

Answer 8

I think now u can do it, can't u?:)

Answer 9

\[\Huge{\color{orange}{a=-10,b=15,c=9}}\]

Answer 10

well now im confused..lol

Answer 11

so itd be( -15 - sq root of 15^2 -4(10*9) )/2(10)

Answer 12

First we took \[t^3\]as common.
Then only u have to factorise the eq in the brackets by the quadratic formula I & @saljudieh07 have given:)

Answer 13

Clear now??

Answer 14

*Using the quadratic formula -is not- factoring.
Using the quadratic formula is wrong here.

Answer 15

yeh i think cliff is right, because the quadratic formula would find the values of t, if the quadratic equation is equal to zero.

so in this case the final answer is:

\[t^3(−10t2+15t+9)\]

Answer 16

Taking a*c = (-10)(9) = -90; there are no integer factors of -90 that add or subtract to make b=15, so -10t^2+15t+9 is prime.

Answer 17

ok..ill go with that

Answer 18

Even...this confused me.. >.>