Polynomial Function given roots and a point
$begingroup$
A 3rd degree polynomial has roots at x=-2i and x=5. The y-intercept is (0,25). Write an equation for this function in factored form with real coefficients.
My first hack is to say that in factored form this is (x-5)(X2+4) as that gets me the roots, but that has y-intercept (0,-20). What am I missing?
Thanks!
polynomials
asked Dec 14 '18 at 2:02
jerHjerH
1104
$endgroup$
add a comment |
$begingroup$
A 3rd degree polynomial has roots at x=-2i and x=5. The y-intercept is (0,25). Write an equation for this function in factored form with real coefficients.
My first hack is to say that in factored form this is (x-5)(X2+4) as that gets me the roots, but that has y-intercept (0,-20). What am I missing?
Thanks!
polynomials
asked Dec 14 '18 at 2:02
jerHjerH
1104
$endgroup$
$begingroup$
Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
$endgroup$
– jerH
Dec 14 '18 at 2:11
$begingroup$
The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
$endgroup$
– SmileyCraft
Dec 14 '18 at 2:15
add a comment |
$begingroup$
A 3rd degree polynomial has roots at x=-2i and x=5. The y-intercept is (0,25). Write an equation for this function in factored form with real coefficients.
My first hack is to say that in factored form this is (x-5)(X2+4) as that gets me the roots, but that has y-intercept (0,-20). What am I missing?
Thanks!
polynomials
asked Dec 14 '18 at 2:02
jerHjerH
1104
$endgroup$
A 3rd degree polynomial has roots at x=-2i and x=5. The y-intercept is (0,25). Write an equation for this function in factored form with real coefficients.
My first hack is to say that in factored form this is (x-5)(X2+4) as that gets me the roots, but that has y-intercept (0,-20). What am I missing?
Thanks!
polynomials
polynomials
asked Dec 14 '18 at 2:02
jerHjerH
1104
asked Dec 14 '18 at 2:02
jerHjerH
1104
asked Dec 14 '18 at 2:02
jerHjerH
1104
asked Dec 14 '18 at 2:02
jerHjerH
1104
asked Dec 14 '18 at 2:02
jerHjerH
1104
1104
$begingroup$
Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
$endgroup$
– jerH
Dec 14 '18 at 2:11
$begingroup$
The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
$endgroup$
– SmileyCraft
Dec 14 '18 at 2:15
add a comment |
$begingroup$
Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
$endgroup$
– jerH
Dec 14 '18 at 2:11
$begingroup$
The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
$endgroup$
– SmileyCraft
Dec 14 '18 at 2:15
$begingroup$
Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
$endgroup$
– jerH
Dec 14 '18 at 2:11
$begingroup$
Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
$endgroup$
– jerH
Dec 14 '18 at 2:11
$begingroup$
The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
$endgroup$
– SmileyCraft
Dec 14 '18 at 2:15
$begingroup$
The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
$endgroup$
– SmileyCraft
Dec 14 '18 at 2:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...
The Roots:
On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as
$$f(x) = a(x - 5)(x - 2i)(x + 2i)$$
up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.
Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).
Finding $a$:
So now you know that the polynomial has the form
$$f(x) = a(x-5)(x^2 + 4)$$
The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.
(Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)
We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.
Plugging in these values, then, we see
$$25 = a(-5)(4) ;;; Rightarrow ;;; 25 = a(-20) ;;; Rightarrow ;;; a = frac{25}{-20} = -frac{5}{4}$$
Thus, the polynomial in its factored form is
$$f(x) = left( -frac{5}{4} right)(x-5)(x^2 + 4)$$
answered Dec 14 '18 at 2:30
Eevee TrainerEevee Trainer
7,02311337
$endgroup$
$begingroup$
In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
$endgroup$
– SmileyCraft
Dec 14 '18 at 3:11
1
$begingroup$
Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 3:13
add a comment |
$begingroup$
You solve the problem by simply multiplying the polynomial by $frac{25}{-20}$.
answered Dec 14 '18 at 2:14
SmileyCraftSmileyCraft
3,631518
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...
The Roots:
On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as
$$f(x) = a(x - 5)(x - 2i)(x + 2i)$$
up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.
Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).
Finding $a$:
So now you know that the polynomial has the form
$$f(x) = a(x-5)(x^2 + 4)$$
The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.
(Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)
We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.
Plugging in these values, then, we see
$$25 = a(-5)(4) ;;; Rightarrow ;;; 25 = a(-20) ;;; Rightarrow ;;; a = frac{25}{-20} = -frac{5}{4}$$
Thus, the polynomial in its factored form is
$$f(x) = left( -frac{5}{4} right)(x-5)(x^2 + 4)$$
answered Dec 14 '18 at 2:30
Eevee TrainerEevee Trainer
7,02311337
$endgroup$
$begingroup$
In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
$endgroup$
– SmileyCraft
Dec 14 '18 at 3:11
1
$begingroup$
Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 3:13
add a comment |
$begingroup$
To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...
The Roots:
On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as
$$f(x) = a(x - 5)(x - 2i)(x + 2i)$$
up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.
Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).
Finding $a$:
So now you know that the polynomial has the form
$$f(x) = a(x-5)(x^2 + 4)$$
The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.
(Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)
We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.
Plugging in these values, then, we see
$$25 = a(-5)(4) ;;; Rightarrow ;;; 25 = a(-20) ;;; Rightarrow ;;; a = frac{25}{-20} = -frac{5}{4}$$
Thus, the polynomial in its factored form is
$$f(x) = left( -frac{5}{4} right)(x-5)(x^2 + 4)$$
answered Dec 14 '18 at 2:30
Eevee TrainerEevee Trainer
7,02311337
$endgroup$
$begingroup$
In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
$endgroup$
– SmileyCraft
Dec 14 '18 at 3:11
1
$begingroup$
Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 3:13
add a comment |
$begingroup$
To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...
The Roots:
On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as
$$f(x) = a(x - 5)(x - 2i)(x + 2i)$$
up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.
Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).
Finding $a$:
So now you know that the polynomial has the form
$$f(x) = a(x-5)(x^2 + 4)$$
The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.
(Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)
We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.
Plugging in these values, then, we see
$$25 = a(-5)(4) ;;; Rightarrow ;;; 25 = a(-20) ;;; Rightarrow ;;; a = frac{25}{-20} = -frac{5}{4}$$
Thus, the polynomial in its factored form is
$$f(x) = left( -frac{5}{4} right)(x-5)(x^2 + 4)$$
answered Dec 14 '18 at 2:30
Eevee TrainerEevee Trainer
7,02311337
$endgroup$
To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...
The Roots:
On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as
$$f(x) = a(x - 5)(x - 2i)(x + 2i)$$
up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.
Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).
Finding $a$:
So now you know that the polynomial has the form
$$f(x) = a(x-5)(x^2 + 4)$$
The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.
(Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)
We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.
Plugging in these values, then, we see
$$25 = a(-5)(4) ;;; Rightarrow ;;; 25 = a(-20) ;;; Rightarrow ;;; a = frac{25}{-20} = -frac{5}{4}$$
Thus, the polynomial in its factored form is
$$f(x) = left( -frac{5}{4} right)(x-5)(x^2 + 4)$$
answered Dec 14 '18 at 2:30
Eevee TrainerEevee Trainer
7,02311337
answered Dec 14 '18 at 2:30
Eevee TrainerEevee Trainer
7,02311337
answered Dec 14 '18 at 2:30
Eevee TrainerEevee Trainer
7,02311337
answered Dec 14 '18 at 2:30
Eevee TrainerEevee Trainer
7,02311337
7,02311337
$begingroup$
In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
$endgroup$
– SmileyCraft
Dec 14 '18 at 3:11
1
$begingroup$
Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 3:13
add a comment |
$begingroup$
In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
$endgroup$
– SmileyCraft
Dec 14 '18 at 3:11
1
$begingroup$
Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 3:13
$begingroup$
In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
$endgroup$
– SmileyCraft
Dec 14 '18 at 3:11
$begingroup$
In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
$endgroup$
– SmileyCraft
Dec 14 '18 at 3:11
1
1
$begingroup$
Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 3:13
$begingroup$
Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 3:13
add a comment |
$begingroup$
You solve the problem by simply multiplying the polynomial by $frac{25}{-20}$.
answered Dec 14 '18 at 2:14
SmileyCraftSmileyCraft
3,631518
$endgroup$
add a comment |
$begingroup$
You solve the problem by simply multiplying the polynomial by $frac{25}{-20}$.
answered Dec 14 '18 at 2:14
SmileyCraftSmileyCraft
3,631518
$endgroup$
add a comment |
$begingroup$
You solve the problem by simply multiplying the polynomial by $frac{25}{-20}$.
answered Dec 14 '18 at 2:14
SmileyCraftSmileyCraft
3,631518
$endgroup$
You solve the problem by simply multiplying the polynomial by $frac{25}{-20}$.
answered Dec 14 '18 at 2:14
SmileyCraftSmileyCraft
3,631518
answered Dec 14 '18 at 2:14
SmileyCraftSmileyCraft
3,631518
answered Dec 14 '18 at 2:14
SmileyCraftSmileyCraft
3,631518
answered Dec 14 '18 at 2:14
SmileyCraftSmileyCraft
3,631518
3,631518
add a comment |
add a comment |
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$begingroup$
Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
$endgroup$
– jerH
Dec 14 '18 at 2:11
$begingroup$
The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
$endgroup$
– SmileyCraft
Dec 14 '18 at 2:15