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.
4c^2-12c+9 is a trinomial because it is in the form of ax^2 + bx + c. In this case, the trinomial is a perfect square trinomial. 4c^2-12c+9 =(2c-3)(2c-3) =(2c-3)^2
No.
No.
-x2 + 2x + 48 = (x +6)(8 - x)
3x2 + 36x + 81 = 3(x2 + 13x + 27)
It is: (x2 + 5x) (x + 8)
(2x+7)(x+7)
x(x + 9)(x + 5)
4c^2-12c+9 is a trinomial because it is in the form of ax^2 + bx + c. In this case, the trinomial is a perfect square trinomial. 4c^2-12c+9 =(2c-3)(2c-3) =(2c-3)^2
No.
No.
(-4x + 3)(x - 2)Apex :)!!
x(x + 4)(x + 6)
(x + 1)(x + 5)
-x2 + 2x + 48 = (x +6)(8 - x)
3x2 + 36x + 81 = 3(x2 + 13x + 27)
2x2 + 15x + 25 = (2x + 5)(x + 5)