What is the factorization of the trinomial 14x square plus 29x-15?

Answer 1

If you just want the answers, the factors are:

(2x+5)(7x-3)=0

and the roots/x-intercepts/zeros are: x= 3/7 and x= -2.5

If you want to see the work...

ax^2 + bx + c = 0

14x^2 + 29x - 15 = 0

This is type 4 factorization... so you need to find numbers that multiply to give you the product of a and c (a*c), and add to give you b.

so, a*c = (14)*(- 15) = -210, and b = 29

the numbers that would satisfy these requirements are +35, and - 6

so, here's the original equation:

14x^2 + 29x - 15 = 0

you carry down the first and third terms (a and c) and place them in the same places

14x^2 - 15 = 0

you then take the numbers you found, 35, and -6, and stick them with x's in the equation...

14x^2 +35x -6x - 15 = 0

you then factor out the largest factor of the first two terms, which is 7x

7x(2x+5)

and then you repeat for the next two terms, the largest factor is -3

-3(2x+5)

so here's your new equation:

7x(2x+5) -3 (2x+5) = 0

note that the two binomials are the same...

take one of the binomials

(2x+5)... this is one of your factors...

and then the others terms (the coefficients) and put them into one binomial

(7x-3) your other factor...

so, your your two factors are

(2x+5)(7x-3)=0

2x=-5

x= -2.5

7x=3

x=3/7

the x values are your roots/zeros/x-intercepts.