Math / Algebra

QuestionSelect each possible formula for the polynomial described below. The degree is $n=2$ and the zeros are $x=6,-14$ . $(x+6)(x+14)$ $(x+6)(x-14)$ $(x-6)(x+14)$ $6(x+6)(x-14)$ $(x-6)^{2}(x+14)^{2}$ $2(x-6)(x+14)$ $(x-6)(x-14)$

Studdy Solution

STEP 1

What is this asking? Find formulas for a parabola that crosses the x-axis at $x = 6$ and $x = -14$ . Watch out! Remember how the sign of the zeros relates to the factors in the formula.
Don't mix them up!

STEP 2

1. Relate zeros to factors
2. Account for scaling
3. Identify correct formulas

STEP 3

We're given that the zeros of the polynomial are $x = 6$ and $x = -14$ .
This means if we plug in $x = 6$ or $x = -14$ into the formula, the result should be zero.

STEP 4

Remember, a zero of a polynomial is a value of $x$ that makes the polynomial equal to zero.
If $x = a$ is a zero, then $(x - a)$ is a factor.

STEP 5

So, if $x = 6$ is a zero, then $(x - 6)$ must be a factor.
Similarly, if $x = -14$ is a zero, then $(x - (-14)) = (x + 14)$ must be a factor.

STEP 6

We've found that $(x - 6)$ and $(x + 14)$ are factors.
Multiplying them together gives us $(x - 6)(x + 14)$ , which is one possible formula.

STEP 7

But wait!
There could be a constant multiplier.
Any formula like $c \cdot (x - 6)(x + 14)$ , where $c$ is a non-zero constant, will still have the same zeros.
This is because if we plug in $x = 6$ or $x = -14$ , the $(x-6)$ or $(x+14)$ factor will become zero, making the entire expression zero, regardless of what $c$ is (as long as $c$ isn't zero itself!).

STEP 8

Let's look at the options and see which ones fit our pattern of $c \cdot (x - 6)(x + 14)$ .

STEP 9

$(x + 6)(x + 14)$ - Nope, wrong signs! $(x + 6)(x - 14)$ - Nope, still wrong signs! $(x - 6)(x + 14)$ - Yes!
This matches our pattern with $c = 1$ . $6(x + 6)(x - 14)$ - Nope, those signs are tricking us again! $(x - 6)^2(x + 14)^2$ - While technically having the correct zeros, the degree of this polynomial is $n=4$ , not $n=2$ as required. $2(x - 6)(x + 14)$ - Yes!
This matches our pattern with $c = 2$ . $(x - 6)(x - 14)$ - So close, but that second sign isn't quite right.

STEP 10

The possible formulas are $(x - 6)(x + 14)$ and $2(x - 6)(x + 14)$ .

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QuestionSelect each possible formula for the polynomial described below. The degree is $n=2$ and the zeros are $x=6,-14$ . $(x+6)(x+14)$ $(x+6)(x-14)$ $(x-6)(x+14)$ $6(x+6)(x-14)$ $(x-6)^{2}(x+14)^{2}$ $2(x-6)(x+14)$ $(x-6)(x-14)$

Studdy Solution

STEP 1

What is this asking? Find formulas for a parabola that crosses the x-axis at $x = 6$ and $x = -14$ . Watch out! Remember how the sign of the zeros relates to the factors in the formula.
Don't mix them up!

STEP 2

1. Relate zeros to factors
2. Account for scaling
3. Identify correct formulas

STEP 3

We're given that the zeros of the polynomial are $x = 6$ and $x = -14$ .
This means if we plug in $x = 6$ or $x = -14$ into the formula, the result should be zero.

STEP 4

Remember, a zero of a polynomial is a value of $x$ that makes the polynomial equal to zero.
If $x = a$ is a zero, then $(x - a)$ is a factor.

STEP 5

So, if $x = 6$ is a zero, then $(x - 6)$ must be a factor.
Similarly, if $x = -14$ is a zero, then $(x - (-14)) = (x + 14)$ must be a factor.

STEP 6

We've found that $(x - 6)$ and $(x + 14)$ are factors.
Multiplying them together gives us $(x - 6)(x + 14)$ , which is one possible formula.

STEP 7

STEP 8

Let's look at the options and see which ones fit our pattern of $c \cdot (x - 6)(x + 14)$ .

STEP 9

STEP 10

The possible formulas are $(x - 6)(x + 14)$ and $2(x - 6)(x + 14)$ .

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Studdy Solution

STEP 1

What is this asking? Find formulas for a parabola that crosses the x-axis at x=6x = 6x=6 and x=−14x = -14x=−14. Watch out! Remember how the sign of the zeros relates to the factors in the formula.Don't mix them up!

STEP 2

1. Relate zeros to factors2. Account for scaling3. Identify correct formulas

STEP 3

We're given that the zeros of the polynomial are x=6x = 6x=6 and x=−14x = -14x=−14.This means if we plug in x=6x = 6x=6 or x=−14x = -14x=−14 into the formula, the result should be zero.

STEP 4

Remember, a *zero* of a polynomial is a value of xxx that makes the polynomial equal to zero.If x=ax = ax=a is a zero, then (x−a)(x - a)(x−a) is a factor.

STEP 5

So, if x=6x = 6x=6 is a zero, then (x−6)(x - 6)(x−6) must be a factor.Similarly, if x=−14x = -14x=−14 is a zero, then (x−(−14))=(x+14)(x - (-14)) = (x + 14)(x−(−14))=(x+14) must be a factor.

STEP 6

We've found that (x−6)(x - 6)(x−6) and (x+14)(x + 14)(x+14) are factors.Multiplying them together gives us (x−6)(x+14)(x - 6)(x + 14)(x−6)(x+14), which is one possible formula.

STEP 7

STEP 8

Let's look at the options and see which ones fit our pattern of c⋅(x−6)(x+14)c \cdot (x - 6)(x + 14)c⋅(x−6)(x+14).

STEP 9

STEP 10

The possible formulas are (x−6)(x+14)(x - 6)(x + 14)(x−6)(x+14) and 2(x−6)(x+14)2(x - 6)(x + 14)2(x−6)(x+14).

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? Find formulas for a parabola that crosses the x-axis at $x = 6$ and $x = -14$ . Watch out! Remember how the sign of the zeros relates to the factors in the formula.
Don't mix them up!

1. Relate zeros to factors
2. Account for scaling
3. Identify correct formulas

We're given that the zeros of the polynomial are $x = 6$ and $x = -14$ .
This means if we plug in $x = 6$ or $x = -14$ into the formula, the result should be zero.

Remember, a zero of a polynomial is a value of $x$ that makes the polynomial equal to zero.
If $x = a$ is a zero, then $(x - a)$ is a factor.

So, if $x = 6$ is a zero, then $(x - 6)$ must be a factor.
Similarly, if $x = -14$ is a zero, then $(x - (-14)) = (x + 14)$ must be a factor.

We've found that $(x - 6)$ and $(x + 14)$ are factors.
Multiplying them together gives us $(x - 6)(x + 14)$ , which is one possible formula.

Let's look at the options and see which ones fit our pattern of $c \cdot (x - 6)(x + 14)$ .

The possible formulas are $(x - 6)(x + 14)$ and $2(x - 6)(x + 14)$ .