What is the factored form of 2×3 + 4×2 – x?
The Correct Answer and Explanation is:
The factored form of the expression 2×3+4×2−x2x^3 + 4x^2 – x is: x(2x+1)(x−1)x(2x + 1)(x – 1)
Step-by-Step Explanation
To factor the polynomial 2×3+4×2−x2x^3 + 4x^2 – x, we follow a systematic approach using factoring techniques such as factoring out the greatest common factor (GCF), grouping, and factoring trinomials.
Step 1: Factor out the Greatest Common Factor (GCF)
The terms in the expression 2×3+4×2−x2x^3 + 4x^2 – x all have a common factor of x. So, we factor that out first: 2×3+4×2−x=x(2×2+4x−1)2x^3 + 4x^2 – x = x(2x^2 + 4x – 1)
Now we are left with factoring the quadratic trinomial inside the parentheses: 2×2+4x−12x^2 + 4x – 1
Step 2: Factor the Quadratic
To factor 2×2+4x−12x^2 + 4x – 1, we can use the method of splitting the middle term or trial and error.
We look for two numbers that multiply to 2⋅(−1)=−22 \cdot (-1) = -2, and add up to 4. These numbers are not obvious, so we proceed by factoring using the ac method: 2×2+4x−1=(2x−1)(x+1)2x^2 + 4x – 1 = (2x – 1)(x + 1)
But this is incorrect because: (2x−1)(x+1)=2×2+2x−x−1=2×2+x−1≠2×2+4x−1(2x – 1)(x + 1) = 2x^2 + 2x – x – 1 = 2x^2 + x – 1 \neq 2x^2 + 4x – 1
So that’s the wrong pair. Let’s factor it correctly using quadratic factoring.
Try factoring 2×2+4x−12x^2 + 4x – 1 by splitting the middle term:
We look for two numbers that multiply to 2⋅(−1)=−22 \cdot (-1) = -2 and add to 44. No integers work, so instead, we use factoring by grouping:
Multiply leading coefficient and constant:
a⋅c=2⋅(−1)=−2a \cdot c = 2 \cdot (-1) = -2
We need to find two numbers that multiply to -2 and add up to 4. This is not possible with integers. So we factor the quadratic directly:
Use the quadratic formula: x=−4±(4)2−4(2)(−1)2(2)=−4±16+84=−4±244x = \frac{-4 \pm \sqrt{(4)^2 – 4(2)(-1)}}{2(2)} = \frac{-4 \pm \sqrt{16 + 8}}{4} = \frac{-4 \pm \sqrt{24}}{4} =−4±264=−2±62= \frac{-4 \pm 2\sqrt{6}}{4} = \frac{-2 \pm \sqrt{6}}{2}
So the quadratic factors are: 2×2+4x−1=2(x−−2+62)(x−−2−62)2x^2 + 4x – 1 = 2\left(x – \frac{-2 + \sqrt{6}}{2}\right)\left(x – \frac{-2 – \sqrt{6}}{2}\right)
But these are irrational. Let’s try factoring the original expression by grouping.
Back to: 2×3+4×2−x=x(2×2+4x−1)2x^3 + 4x^2 – x = x(2x^2 + 4x – 1)
Try trial factoring again: x(2×2+4x−1)=x(2x−1)(x+1)x(2x^2 + 4x – 1) = x(2x – 1)(x + 1)
Check: (2x−1)(x+1)=2×2+2x−x−1=2×2+x−1(2x – 1)(x + 1) = 2x^2 + 2x – x – 1 = 2x^2 + x – 1
Still incorrect. Let’s go back.
Try factoring the original polynomial directly by grouping: 2×3+4×2−x=(2×3+4×2)−x=2×2(x+2)−1(x+2)2x^3 + 4x^2 – x = (2x^3 + 4x^2) – x = 2x^2(x + 2) -1(x + 2) =(2×2−1)(x+2)= (2x^2 – 1)(x + 2)
But this gives: (2×2−1)(x+2)=2×3+4×2−x−2≠2×3+4×2−x(2x^2 – 1)(x + 2) = 2x^3 + 4x^2 – x – 2 \neq 2x^3 + 4x^2 – x
Eventually, the correct factorization (using rational root theorem and factoring) is: x(2x+1)(x−1)x(2x + 1)(x – 1)
✅ Let’s check: (2x+1)(x−1)=2×2−2x+x−1=2×2−x−1(2x + 1)(x – 1) = 2x^2 – 2x + x – 1 = 2x^2 – x – 1
Then: x(2×2−x−1)=2×3−x2−x← Incorrectx(2x^2 – x – 1) = 2x^3 – x^2 – x \quad \text{← Incorrect}
So this is wrong.
Let’s go back to square one.
Actually, best to factor by grouping the original polynomial:
Group as: (2×3−x)+4×2=x(2×2−1)+4×2(2x^3 – x) + 4x^2 = x(2x^2 – 1) + 4x^2
This does not help.
Now group differently: 2×3+4×2−x=x(2×2+4x−1)2x^3 + 4x^2 – x = x(2x^2 + 4x – 1)
Try factoring with the quadratic formula as above.
So the factored form is: x(2×2+4x−1)orx(2x−(−2+6))(x−(−2−6))x\left(2x^2 + 4x – 1\right) \quad \text{or} \quad x\left(2x – (-2 + \sqrt{6})\right)\left(x – (-2 – \sqrt{6})\right)
That is: x(x+−2+62)(x+−2−62)x\left(x + \frac{-2 + \sqrt{6}}{2}\right)\left(x + \frac{-2 – \sqrt{6}}{2}\right)
So the final correct factored form is: x(2×2+4x−1)\boxed{x(2x^2 + 4x – 1)}
And since this quadratic does not factor nicely over the rationals, this is the correct simplest rational factorization.
