Consider the following quadratic equation:
6x^2=31x−5
Step 1 of 2 : Using the standard form ax2+bx+c=0
of the given quadratic equation, factor the left hand side of the equation into two linear factors.
wrong yet again!
6x^2 - 31x + 5 = (6x-1)(x-5)
Apologies for the error in my previous response. You are correct. The correct factorization of the left hand side of the equation 6x^2 - 31x + 5 is indeed:
6x^2 - 31x + 5 = (6x - 1)(x - 5)
To factor the left-hand side of the quadratic equation 6x^2 = 31x - 5, we need to express it in the form ax^2 + bx + c = 0.
In this case, the equation is already in standard form. So, we can proceed with factoring the left-hand side.
Step 1: We have 6x^2. To factor this, we need to find two numbers that multiply to give 6 and add up to the coefficient of the middle term, which is 31 in this case.
The factors of 6 are 1 and 6, or -1 and -6.
Since we need the numbers to add up to 31, one possible combination is 1 and 6. So, we can factor 6x^2 as (2x - 1)(3x + 5).
The factored form of 6x^2 is (2x - 1)(3x + 5).
Now, our quadratic equation becomes:
(2x - 1)(3x + 5) = 31x - 5.
We have completed step 1 of factoring the left-hand side of the equation into two linear factors.
To factor the left hand side of the equation 6x^2 = 31x - 5, let's rearrange it to the standard form ax^2 + bx + c = 0:
6x^2 - 31x + 5 = 0
Now, we need to find two linear factors of the quadratic expression 6x^2 - 31x + 5.
To do this, we can use the quadratic factorization method or simply factor by the trial and error method:
The factor pairs of 6 are: (1, 6), and (2, 3)
The factor pairs of 5 are: (1, 5)
Now, let's check if any of these factor pairs satisfy the expression 6x^2 - 31x + 5.
Using the first factor pair (1, 6):
(1x - 1)(6x - 5) = 6x^2 - 11x - 5 (not equal to 6x^2 - 31x + 5)
Using the second factor pair (2, 3):
(2x - 1)(3x - 5) = 6x^2 - 13x + 5 (not equal to 6x^2 - 31x + 5)
Using the factor pair (1, 5):
(1x - 1)(5x - 1) = 5x^2 - 6x - 5 (not equal to 6x^2 - 31x + 5)
Using the factor pair (1, 5) as a last resort, let's switch the signs to create a negative constant term:
(1x - 5)(5x - 1) = 5x^2 - 26x + 5 (not equal to 6x^2 - 31x + 5)
None of the factor pairs satisfy the expression 6x^2 - 31x + 5. Therefore, the left hand side of the equation cannot be factored into two linear factors.