How would you factor 8x^2+14x+5?

Question

How would you factor 8x^2+14x+5?

Answer 1

really hard to explain over computer but the factors are (2x+1)(4x+5). let me attempt an explanation:

first you have to find two factors of ac that when added equal b. so find two factors of 8x5 [40] that when added together give you 14. the factors are 4 and 10. keep that in mind.

now the first numbers of your factors when multiplied must equal 8x^2 and the second numbers when multiplied must equal 5. we are going to use 2x and 4x for the first numbers:

(2x+__)(4x+__)

the only factors of 5 are 1 and 5. but which number goes where?

recall the 4 and 10 that we talked about earlier? if we put 5 in the first blank and 1 in the second blank we get:

(2x+5)(4x+1)

when multiplied we get 8x^2 +2x+20x+5. the 2 and 20 is not what we wanted. we want 4 and 10.

so let's try (2x+1)(4x+5):

8x^2 +10x+4x+5 perfect.

I hope I didn't confuse you

Answer 2

(2x+1)(4x+5)

Answer 3

Maybe try use the quadratic formula:

[-b ± √(b² - 4ac)] / 2a

[-14 ± √(14² - 4*8*5)] / 2*8

[-14 ± √(196 - 160)] / 16

(-14 ± √36) / 16

(-14 ± 6) / 16

Bring out 2 as a common factor:

2(-7 ± 3) / 16

2 cancel out:

(-7 ± 3) / 8

-10/8 = -5/4

-4 / 8 = -1/2

Therefore:

(x + 5/4)(x + 1/2)

Just simplify it a bit:

(4x + 5)(2x + 1)

Answer 4

8x^2+10x+4x+5=0

2x(4x+5)+1(4x+5)=0

(2x+1)(4x+5)=0

x=-1/2,-5/4

Answer 5

(4x + 5 ) (2x + 1)

Good luck to you !

Answer 6

The first numbers in each parentheses have to multiply to give 8x^2

could be 8x and x  .... or... 2x and 4x

The last numbers in each parentheses have to multiply to give 5

could be 1 and 5  .... or.... -1 and -5

From there you have to try different combinations to see which is right...

(2x + 1)(4x + 5)

And my TI-89 agrees with me.

Take care,

David