Factor each polynomial. 7(3k-1)^{2}+26(3k-1)-8

veksetz

veksetz

Answered question

2021-12-09

Factor each polynomial.
7(3k1)2+26(3k1)8

Answer & Explanation

Foreckije

Foreckije

Beginner2021-12-10Added 32 answers

Step 1
Given:
7(3k1)2+26(3k1)8
A polynomial is factored completely when it is written as product of prime polynomials.
FOIL METHOD :
(Ax+B)(Cx+D)=ACx2+(BC+AD)x+BD.Ax+BCx+D
To factorize 7(3k1)2+26(3k1)8
Substitute 3k-1 = x
The equation becomes, 7x2+26x8
Using Foil method, AC=7 and BD=-8
The positive factors of 7 are 1 and 7.
Because the middle term has the positive coefficient and the last only the combination of term has a negative coefficient, we consider negative and positive factors of -8.
The possibilities of -8 are 1 and -8 or -1 and 8 or 2 and -4 or 4 and -2.
Step 2
Now we try various arrangements of these factors until we find one that gives the correct coefficient of x.
We first take the possibilities 1 and 7 and 1 and -8 and check the answer,
(x+7)(x8)=x2+7x8x56
=x2x56
Answer is incorrect.
We take the possibilities 7 and 1 and 1 and -8 and check :
(7x+1)(x8)=7x256x+x87x
=7x255x87
Answer is incorrect
We take the possibilities 7 and 1 , and -2 and 4 and check :
(7x2)(x+4)=7x2+28x2x87
=7x2+26x87
Answer is correct.
So, the factor of 7x2+26x87 is (7x-2)(x+4)
Step 3
Now, again substitute 3k-1 = x
7(3k1)2+26(3k1)8
=(7(3k-1)-2)(3k-1+4)
=(21k-7-2)(3k+3)
=(21k-9)(3k+3)
Take out common term outside,
21k-9=3(7k-3)
3k+3=3(k+1)
(21k-9)(3k+3)=3(7k-3)3(k+1)
=9(7k-3)(k+1)
Therefore, the factor of 7(3k1)2+26(3k1)8 is 9(7k-3)(k+1)

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