`7u^2 - 4u - 3` Factor the trinomial. If the trinomial cannot be factored, say so.

Textbook Question

Chapter 5, 5.2 - Problem 44 - McDougal Littell Algebra 2 (1st Edition, Ron Larson).
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atyourservice's profile pic

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

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We can solve this using the AC factoring method:

`7u^2-4u-3`

Multiply a by c

`7 xx -3 = -21`

Find factors of -21 that add up to b (-4)

these numbers will be 3 and -7, plug these in as b

`7u^2-7u + 3u -3`

now group:

`(7u^2-7u) + (3u -3)`

Factor out the greatest common factors:

7u (u - 1 ) + 3 (u - 1 )

Put the numbers outside in a parentheses together :

(7u + 3)(u - 1)

This is the factored version of the equation.

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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Another way of factoring a trinomial is by equating it to 0 and determining the roots of the quadratic equation using the quadratic formula. The roots of `ax^2 + bx + c = 0` are `(-b+-sqrt(b^2-4ac))/(2a)`

The roots of 7u^2-4u-3 = 0 are:

`(4+-sqrt(16+84))/(14) = (4+-10)/14` or `14/14 = 1` and `-6/14 = -3/7`

The trinomial can be written as (u - 1)(u+3/7)

This can be written in the form (u-a)(u-b) where a and b are integers as:

(u -1)(7u+3)

The factorized form of 7u^2-4u-3 is (u -1)(7u+3)

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udonbutterfly | Student, College Freshman | (Level 1) Valedictorian

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Okay Lets start with multiply 7 by 3 which will equal...

7X3=21

Now we have to have multiples of 21 that when add/subtracted will gives us the middle term so -4u

1x21 21-1 this will not give us -4

7X3 3-7 this will gives us -4! Now replace these two numbers with the middle term

7u^2-4u-3

(7u^2-7u)+(3u-3)

Now take out the number that each term is divisible by

7u(u-1)+3(u-1) The expressions in the parentheses should match because it will only be one expression for the final equation. The second expression will be the combination of the two terms you got when you took out the number the expressions were divisible by

(7u+3)(u-1) And here's the answer!

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