1. Factor a^2 + 28a + 27.?

Hi Nikki...

The surest way to find the factors of your problems is to use the quadratic formula.. It will ease the burden of lengthy "trial and error" method.

The standard quadratic form of a trinomial is ax^2 + bx + c = 0
where;
a = coefficient of x^2
b = coefficient of x
c = constant

To get the two (2) values of x, x(-) and (x-) symbolized as x(±), you must substitute these coefficients and constant to the Quadratic formula;

x(±) = [- b ± (b^2 - 4ac)^(1/2)] /2a ---->>> QUADRATIC FORMULA

1. Given: a^2 + 28a + 27

Required: Factor:

Solution:

The standard form the given equation is a^2 + 28a + 27 = 0

Coefficients of a and constant:
A = 1
B = 28
C = 27

a(±) = [- B ± (B^2 - 4AC)^(1/2)] /2A
a(±) = {- 28 ± [(28)^2 - 4(1)(27)] }/2(1)
a(±) = {- 28 ± √676}/2
a(±) = {- 28 ± 26}/2

a(+) = (- 28 + 26)/2
a(+) = -2/2 = - 1 ------------>>> value of a(+)

a(-) = (- 28 - 26)/2
a(-) = - 54/2 = - 27 --------->>>> value of a(-)

thus, the factor of a^2 + 28a + 27 is;

(a + 1) (a + 27) -------->>>> answer

2. Given: (4p - 3)(5p + 1)

Required: Product

Solution:

(4p - 3)(5p + 1) = 20p^2 - 11p - 3 ------>>> Ans

3. Given: 2x^2 - 9x - 11

Required: Factor

Solution:

The standard form of the equation is;
2x^2 - 9x - 11 = 0

Coefficients of x and constant
a = 2
b = - 9
c = - 11

Substitute these to the the QUADRATIC FORMULA once again,

x(±) = [- b ± (b^2 - 4ac)]/2a
x(±) = {- ( -9) ± (- 9)^2 - 4(2)(-11)]^(1/2)}/2(2)
x(±) = {9 ± √169}/4
x(±) = {9 ± 13}/4

x(+) = (9 + 13)/4
x(+) = 11/2 ----------------->>> value of x(+)

x(-) = (9 - 13)/4
x(-) = - 1 --------------------->>> value of x(-)

thus, the factor of the given equation 2x^2 - 9x - 11 is

(x - 11/2) (x + 1)

or

(2x - 11) ( x + 1) ------------------->>> ans

4. Given: 8n^2 - 26n + 15

Required: Factor

solution:

The standard form of the given equation is;

8n^2 - 26n + 15 = 0

Coefficients of n and constant:
a = 8
b = - 26
c = 15

Substitute these to the QUADRATIC FORMULA;

n(±) = [- b ± (b^2 - 4ac)^(1/2)] /2a
n(±) = {- (-26) ± [(-26)^2 - 4(8)(15)^](1/2)} /2(8)
n(±) = {26 ± √196}/16
n(±) = {26 ± 14}/16

n(+) = (26 + 14)/16
n(+) = 5/2 ------------------>>>> value of n(+)

n(-) = (26 - 14)/16
n(-) = 12/16
n(-) = 3/4 ------------------>>>> value of n(-)

thus, the factor of 8n^2 - 26n + 15 is;

(n - 5/2) (n - 3/4)
or
(2n - 5) (4n - 3) ------------------->>>> ans

5. Given: 81k^2 + 36k + 4

Required: Factor

Solution:

The standard form of the given equation is;
81k^2 + 36k + 4 = 0

Coefficients of k and constant;
a = 81
b = 36
c = 4

Substitute these to the QUADRATIC FORMULA;
k(±) = [- b ± (b^2 - 4ac)^(1/2)] /2a
k(±) = {- 36 ± [(36)^2 - 4(81)(4)]} /2(81)
k(±) = {- 36 ± 0}/2(81)

k(+) = - 36/2(81)
k(+) = - 2/9 ----------------->>> value of k(+)

k(-) = - 2/9 ----------------->>> value of k(-)

thus, the factor of 81k^2 + 36k + 4 is;

(k + 2/9) (k + 2/9)

or

(9k + 2) (9k + 2) --------->>> ans

6. Given: (2y + 3)(2y - 3)

Required: find the Product

Solution:

(2y + 3)(2y - 3)
= 4y^2 - 9 -------------------->>>> ans

1. (a + 27) (a + 1)

Factor a^2 + 28a + 27.?


Answer: (a+14+9)(a+14+3)

Rank me as best answer cuz i helped!~~!!!!

First, take out a common factor of 3: 3{(a - 1)^2 - 9}. The term (a - 1)^2 - 9 can be factorised as the difference of two squares: a^2 - b^2 = (a + b)(a - b): 3{(a - 1)^2 - 9} = 3{(a - 1) + 3}{(a - 1) - 3} = 3(a + 2)(a - 4). Hope this helps. Twiggy.

1. a^2+28a+27 = (a+27)(a+1)
2. (4p-3)(5p+1) = 20p^2+4p-15p-3
=20p^2-11p-3
3. 2x^2-9x-11 =(2x-11)(x+1)
4. 8n^2-26n+15 = (4x-5)(2x-3)
5. 81k^2+36k+4 =(9x+2)^2
6. (2y+3)(2y-3) = 4y^2-9

1)a^2+27a+a+27
a(a+27)+1(a+27)
(a+1)(a+27)

3) 2x^2-9x-11
2x^2+2x-11x-11
2x(x+1)-11(x+1)
(x+1)(2x-11)

5) 81k^2+18k+18k+4
9k(9k+2)+2(9k+2)
(9k+2)(9k+2)

1)
a^2 + 28a + 27
= (a + 27)(a + 1)

2)
(4p - 3)(5p + 1)
= 20p^2 - 15p + 4p - 3
= 20p^2 - 11p - 3

3)
2x^2 - 9x - 11
= (2x - 11)(x + 1)

4)
8n^2 - 26n + 15
= (4n - 3)(2n - 5)

5)
81k^2 + 36k + 4
= (9k + 2)(9k + 2)

6)
(2y + 3)(2y - 3)
= 4y^2 + 6y - 6y - 9
= 4y^2 - 9

1) a^2 + 28a + 27
a^2 +a+27a + 27
a(a+1)+27(a + 1)
(a+1)(a+27)

2)(4p - 3)(5p + 1)
4p(5p + 1)-3(5p+1)
20p^2 + 4p-15p-3
20p^2 -11p-3

3) 2x^2 - 9x - 11
2x^2 +2x- 11x - 11
2x(x +1)- 11(x +1)
(2x-11)(x+1)

4) 8n^2 - 26n + 15.
8n^2 - 20n -6n+ 15.
4n(2n - 5) -3(2n-5)
(4n-3)(2n-5)

5) 81k^2 + 36k + 4.
81k^2 + 36k + 4.
81k^2 + 18k+18k + 4.
9k(9k +2)+2(9k+2)
(9k +2)(9k+2)


6) (2y + 3)(2y - 3)
this is of the form (a+b)(a-b)
so (a+b)(a-b)=a^2-b^2
here a=2y & b=3

so substituting in (a+b)(a-b)=a^2-b^2
we get
(2y+3)(2y-3)=(2y)^2-(3)^2
(2y+3)(2y-3)=4y^2-9

1. (a+27) (a+1)
2. 20p^2-11p-3
3.(2x_11) (x+1)
4.(4n-3) (2n-5)
5.(9k+2) (9k+2)
6. 4y^2-9

do your own homework so you don't end up with a McJob.