Question

Help! Is the binomial a factor of the polynomial function? f(x)=2x^3+15x^2+22x-15.

Answer 1

Answer:

Yes − (2x - 1) → $\displaystyle (x + 3)(x + 5)$

No − (2x + 1) → $\displaystyle x^2 + 7x + 7\frac{1}{2} - \frac{45}{2[2x + 1]}$

No − (x - 5) → $\displaystyle 2x^2 + 25x + 147 + \frac{720}{x - 5}$

Yes − (x + 5) → $(x + 3)(2x - 1)$

No − (x - 3) → $\displaystyle 2x^2 + 21x + 85 + \frac{240}{x - 3}$

Yes − (x + 3) → $(2x - 1)(x + 5)$

Step-by-step explanation:

The binomials of $\displaystyle (2x - 1)\:and\:(2x + 1),$are not in the form of x - c, so we have to use long polynomial division:

2x + 1

$\displaystyle x^2 = \frac{2x^3}{2x}$

x²[2x + 1] = 2x³ + x²

$\displaystyle -[2x^3 + x^2] + [2x^3 + 15x^2 + 22x - 15] = 14x^2 + 22x - 15$

$\displaystyle x^2 + \frac{14x^2 + 22x - 15}{2x + 1}$

$\displaystyle 7x = \frac{14x^2}{2x}$

7x[2x + 1] = 14x² + 7x

$\displaystyle -[14x^2 + 7x] + [14x^2 + 22x - 15] = 15x - 15$

$\displaystyle x^2 + 7x + \frac{15x - 15}{2x + 1}$

$\displaystyle 7\frac{1}{2} = \frac{15}{2}$

7½[2x + 1] = 15x + 7½

$\displaystyle -[15x + 7\frac{1}{2}] + [15x - 15] = -22\frac{1}{2}$

$\displaystyle x^2 + 7x + 7\frac{1}{2} - \frac{22\frac{1}{2}}{2x + 1}$

$\displaystyle x^2 + 7x + 7\frac{1}{2} - \frac{45}{2[2x + 1]}$ → final answer

Since this quotient has a remainder of $\displaystyle -\frac{45}{2[2x + 1]},$it is NOT a factor of $\displaystyle 2x^3 + 15x^2 + 22x - 15.$

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2x - 1

By the Rational Root Theorem, we take the Least Common Divisor [LCD] between the leading coefficient of 2, and the initial value of −15, which is 1, but in this case, since −15 is a number we can work with, we take −3 [15 is negative, so 3 has to be negative as well], so this automatically makes ANOTHER factor of $x + 3$. Next, since the factor\divisor is in the form of $x - c$, use what is called Synthetic Division. Remember, in this formula, −c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:

−3| 2 15 22 −15

↓ −6 −27 15

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2 9 −5 0 → $\displaystyle 2x^2 + 9x - 5$

You start by placing the c in the top left corner, then list all the coefficients of your dividend [2x³ - 15x² + 22x - 15]. You bring down the original term closest to c then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that 2 in your quotient can be a 2x², the 9x follows right behind it, and bringing up the rear, −5, giving you the quotient of $\displaystyle 2x^2 + 9x - 5.$

Finally, you can just simply factor this quotient:

$\displaystyle 2x^2 + 9x - 5\\ \\ [2x^2 - x] + [10x - 5] \\ x[2x - 1] \: \: 5[2x - 1] \\ \\ [x + 5][2x - 1]$

So altogether, so far, we ALREADY KNOW that the factors of $\displaystyle [x + 3], [x + 5], and\:[2x - 1],$ are all factors of $\displaystyle 2x^3 + 15x^2 + 22x - 15,$but we still have to find out if the last two factors of $\displaystyle (x - 5)\:and\:(x - 3)$are factors of the dividend using Synthetic Division again:

x - 5

5| 2 15 22 −15

↓ 10 125 735

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2 25 147 720 → $\displaystyle 2x^2 + 25x + 147 + \frac{720}{x - 5}$

Since this quotient has a remainder of $\displaystyle \frac{720}{x - 5},$it is NOT a factor of $\displaystyle 2x^3 + 15x^2 + 22x - 15.$

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x - 3

3| 2 15 22 −15

↓ 6 63 255

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2 21 85 240 → $\displaystyle 2x^2 + 21x + 85 + \frac{240}{x - 3}$

Since this quotient has a remainder of $\displaystyle -\frac{240}{x - 3},$it is NOT a factor of $\displaystyle 2x^3 + 15x^2 + 22x - 15.$

Well, there you have it. We have determined EVERY binomial as a non-factor and factor.

I am delighted to assist you anytime my friend!

Answer 2

Answer:

yes, no, yes, no, no, yes

Step-by-step explanation:

Check if the zero of each binomial is a zero of the polynomial.  If it is, then the binomial is a factor.

f(-3) = 0

f(3) = 240

f(-5) = 0

f(5) = 720

f(-1/2) = -22.5

f(1/2) = 0