SOLUTION: Which binomial is a factor of: 2x^2 + x - 3? A. (x + 1) B. (2x + 3) C. (2x - 3) D. (2x - 1)

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Which binomial is a factor of: 2x^2 + x - 3? A. (x + 1) B. (2x + 3) C. (2x - 3) D. (2x - 1)      Log On


   

Question 1205772: Which binomial is a factor of:
2x^2 + x - 3?
A. (x + 1)
B. (2x + 3)
C. (2x - 3)
D. (2x - 1)
Found 4 solutions by MathLover1, greenestamps, math_tutor2020, josgarithmetic:
Answer by MathLover1(20850) About Me  (Show Source):
Answer by greenestamps(13216) About Me  (Show Source):
You can put this solution on YOUR website!


The quadratic term in the polynomial is 2x^2, and the absolute value of the constant term is 3. In all the answer choices, the linear term is either x or 2x, and the constant term has an absolute value of either 1 or 3. So any of the answer choices could be right.

You could answer the question by factoring the given quadratic using your favorite method, or by using the quadratic formula to find the two roots.

But with the problem as posed, you can also get some good experience by considering, for each answer choice, what the other factor would have to be to get a quadratic term of 2x^2 and a constant term of -3. Then see if the product is the correct polynomial.

A. (x + 1)
The other factor would have to be (2x - 3), giving a product of 2x^2-x-3 -- not right.

B. (2x + 3)
The other factor would have to be (x-1), giving a product of 2x^2+x-3 -- correct.

ANSWER: The other factor is (2x+3)

We found the right answer, so we don't need to check answer choices C and D.

NOTE: With answer choice A, (x+1), we found the product to be (x+1)(2x-3) = x^2-x-3; the product we wanted is x^2+x-3. The coefficients are the right size, but the sign of the linear term in the product is wrong. When that is the case, the two factors have the right coefficients, but the signs of the two factor are switched. So instead of (x+1)(2x-3) being the factorization, it is (x-1)(2x+3).

So we found answer choice B to be correct by trying it. But in trying answer choice A and finding (x+1)(2x-3) = 2x^2-x-3, where we wanted the product to be 2x^2+x-3 (correct except for the sign of the linear term), we could have concluded at that point that the factorization is (x-1)(2x+3), making answer choice B correct.

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: (2x+3) which is choice B

Explanation

There are a few approaches we could take, as demonstrated by the great solutions by the other tutors.

I'll use the quadratic formula to find the roots.
Then I'll use those roots to construct the factorization.

The equation 2x%5E2%2Bx-3+=+0 is of the form ax%5E2%2Bbx%2Bc+=+0
Notice that: a = 2, b = 1, c = -3

Use the quadratic formula to solve for x.
x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-1%2B-sqrt%28%281%29%5E2-4%282%29%28-3%29%29%29%2F%282%282%29%29

x+=+%28-1%2B-sqrt%281+%2B+24%29%29%2F%284%29

x+=+%28-1%2B-sqrt%2825%29%29%2F%284%29

x+=+%28-1%2B-++5%29%2F%284%29

x+=+%28-1%2B5%29%2F%284%29 or x+=+%28-1-5%29%2F%284%29

x+=+%284%29%2F%284%29 or x+=+%28-6%29%2F%284%29

x+=+1 or x+=+-3%2F2
Feel free to skip a few steps if these seem a bit verbose.

The root x+=+1 leads to x-1+=+0 showing that (x-1) is one factor.

The root x+=+-3%2F2 leads to 2x+=+-3 and ultimately 2x%2B3+=+0.
That makes (2x+3) the other factor.

Ultimately 2x%5E2%2Bx-3 factors to %28x-1%29%282x%2B3%29
You can use the FOIL rule to expand out (x-1)(2x+3) to get the original expression to help verify things.

We can claim that 2x%5E2%2Bx-3=%28x-1%29%282x%2B3%29 is an identity.
An identity is a true equation for all valid x in the domain, which in this case is the set of all real numbers.


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Another approach

For this section I'll use the remainder theorem.
If you prefer the previous section then you can ignore this current section.

The remainder theorem states:
If p(x) divides over (x-k) then p(k) is the remainder.

Consequently an extension is that if (x-k) is a factor of p(x) then p(k) = 0.
The proof is left to the reader.
Hint: p(x) = (x-k)*q(x)

The expression x+1 is same as writing x - (-1) to show that k = -1 here.
Then that value of k is plugged into x
p(x) = 2x^2 + x - 3
p(-1) = 2(-1)^2 + (-1) - 3
p(-1) = -2
The nonzero result tells us that (x+1) is NOT a factor of 2x^2+x-3

The expression 2x+3 appears to not fit the format x - k.
But with a bit of algebraic adjustments we could say:
2x+3 = 2(x+3/2) = 2(x - (-3/2))

Then we just focus on the x - (-3/2) portion inside.
We have k = -3/2, which leads to,
p(x) = 2x^2 + x - 3
p(-3/2) = 2(-3/2)^2 + (-3/2) - 3
p(-3/2) = 0
The result of 0 tells us that (2x+3) is a factor.

You should find choices C and D produce nonzero results.
I'll let the student check those.

Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E2%2Bx-3
might expect -1 or -3 to be one of the constant terms in one of the binomial factors.
(2x - 1)? (x-1)? (2x-3)? (x-3)?

Trying simple factorization of the trinomial
(2x-1 )(x+3 )=...x-6x=-5x
-
(2x-3)(x+1)=...-3x+2x this one worked.
-
(2x+1)(x-3)=no need to try
-
(2x+3)(x-1)=no need to try

The factorization showing as the binomials: (2x-3)(x+1)