Mixing
Factorise the following p(x) as the product of a linear term and a quadratic polynomial with no real roots.
Factorise the following p(x) functions as the product of a linear term and a quadratic polynomial with no real roots (or if there are real roots, factorise to irreducible form and find them):
1. $$p(x)=x^3+x$$
2. $$p(x)=x^4+x^2$$
3. $$p(x)=x^3-x$$
linear-algebra polynomials roots factoring irreducible-polynomials
asked Nov 22 '18 at 4:36
Kayla MartinKayla Martin
636
add a comment |
Factorise the following p(x) functions as the product of a linear term and a quadratic polynomial with no real roots (or if there are real roots, factorise to irreducible form and find them):
1. $$p(x)=x^3+x$$
2. $$p(x)=x^4+x^2$$
3. $$p(x)=x^3-x$$
linear-algebra polynomials roots factoring irreducible-polynomials
asked Nov 22 '18 at 4:36
Kayla MartinKayla Martin
636
What have you tried? Where are you confused?
– Dzoooks
Nov 22 '18 at 4:45
add a comment |
Factorise the following p(x) functions as the product of a linear term and a quadratic polynomial with no real roots (or if there are real roots, factorise to irreducible form and find them):
1. $$p(x)=x^3+x$$
2. $$p(x)=x^4+x^2$$
3. $$p(x)=x^3-x$$
linear-algebra polynomials roots factoring irreducible-polynomials
asked Nov 22 '18 at 4:36
Kayla MartinKayla Martin
636
Factorise the following p(x) functions as the product of a linear term and a quadratic polynomial with no real roots (or if there are real roots, factorise to irreducible form and find them):
1. $$p(x)=x^3+x$$
2. $$p(x)=x^4+x^2$$
3. $$p(x)=x^3-x$$
linear-algebra polynomials roots factoring irreducible-polynomials
linear-algebra polynomials roots factoring irreducible-polynomials
asked Nov 22 '18 at 4:36
Kayla MartinKayla Martin
636
asked Nov 22 '18 at 4:36
Kayla MartinKayla Martin
636
asked Nov 22 '18 at 4:36
Kayla MartinKayla Martin
636
asked Nov 22 '18 at 4:36
Kayla MartinKayla Martin
636
asked Nov 22 '18 at 4:36
Kayla MartinKayla Martin
636
636
What have you tried? Where are you confused?
– Dzoooks
Nov 22 '18 at 4:45
add a comment |
What have you tried? Where are you confused?
– Dzoooks
Nov 22 '18 at 4:45
What have you tried? Where are you confused?
– Dzoooks
Nov 22 '18 at 4:45
What have you tried? Where are you confused?
– Dzoooks
Nov 22 '18 at 4:45
add a comment |
2 Answers
2
active
oldest
votes
Really, these polynomials target a single skill: taking out a common factor. For example, with $x^2+x$, both terms have $x$ as a factor, so you can divide it out (inverse distributive property of sorts) and get $x(x+1)$.
$$$$
$$x^3+x = x(x^2+1)$$
Roots are $0$, $i$, and $-i$.
$$x^4+x^2 = x^2(x^2+1)$$
Roots are $0$, $i$, and $-i$.
$$x^3-x = x(x^2-1) = x(x+1)(x-1)$$
Roots are $0$, $1$, and $-1$.
answered Nov 22 '18 at 4:49
Christopher MarleyChristopher Marley
1,020115
Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
– Kayla Martin
Nov 22 '18 at 4:54
@KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
– idea
Nov 22 '18 at 4:59
Okay, thank you!
– Kayla Martin
Nov 22 '18 at 5:01
add a comment |
I think it's important you understand this idea for more difficult cases than you ask about, so my answer will touch on a more general method for solving these problems, and show you how it works for slightly more difficult cases.
We have that if for some polynomial $p(x)$, $p(a)=0$, then:
$$p(x)=(x-a)cdot q(x)$$
Where $q(x)$ is a polynomial one degree smaller than $p(x)$ (the degree of a polynomial just means the largest power of $x$ in it).
If you can find any point where a polynomial $p(x) $ has a zero, you can take out the factor in this way and find the smaller polynomial $q(x)$, which can either be solved or reduced again.
Let me demonstrate first on (b)
We have that $p(x)=x^4-x^2$
I like to test values $0,1,-1,2,-2$ etc... Here, it's plain that $p (0)=0$, so we take out $(x-0)$, or just $x$. We now have that:
$$x^4-x^2=xcdot q(x)$$
From which we get that $q(x)=x^3-x$. But $q(0)=0$, so we can take out another factor $x$, giving $$x^4-x^2=x^2cdot r (x)$$ It follows that $r(x)=x^2+1$, a quadratic we can solve by the Quadratic Formula.
This is a rather bland example, so I'll demonstrate the procedure on a better example.
Suppose we have $$f(x)=x^3+4x^2+7x+4$$
Plugging in values for $x$, we get that $f(-1)=0$
So we have that:
$$x^3+4x^2+7x+4=(x+1)cdot g(x)$$
Or rather,
$$x^3+4x^2+7x+4=(x+1)(x^2+px+q)$$ for some real $p,q$.
Expanding this out we get:
$$x^3+4x^2+7x+4=x^3+(p+1)x^2+(p+q)x+q$$
From which it follows that $p=3, q=4$.
So: $$f(x)=(x+1)(x^2+3x+4)$$, which is a quadratic we can solve (even if the solutions aren't real)
answered Nov 22 '18 at 5:50
Rhys HughesRhys Hughes
5,0541427
add a comment |
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2 Answers
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2 Answers
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votes
Really, these polynomials target a single skill: taking out a common factor. For example, with $x^2+x$, both terms have $x$ as a factor, so you can divide it out (inverse distributive property of sorts) and get $x(x+1)$.
$$$$
$$x^3+x = x(x^2+1)$$
Roots are $0$, $i$, and $-i$.
$$x^4+x^2 = x^2(x^2+1)$$
Roots are $0$, $i$, and $-i$.
$$x^3-x = x(x^2-1) = x(x+1)(x-1)$$
Roots are $0$, $1$, and $-1$.
answered Nov 22 '18 at 4:49
Christopher MarleyChristopher Marley
1,020115
Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
– Kayla Martin
Nov 22 '18 at 4:54
@KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
– idea
Nov 22 '18 at 4:59
Okay, thank you!
– Kayla Martin
Nov 22 '18 at 5:01
add a comment |
Really, these polynomials target a single skill: taking out a common factor. For example, with $x^2+x$, both terms have $x$ as a factor, so you can divide it out (inverse distributive property of sorts) and get $x(x+1)$.
$$$$
$$x^3+x = x(x^2+1)$$
Roots are $0$, $i$, and $-i$.
$$x^4+x^2 = x^2(x^2+1)$$
Roots are $0$, $i$, and $-i$.
$$x^3-x = x(x^2-1) = x(x+1)(x-1)$$
Roots are $0$, $1$, and $-1$.
answered Nov 22 '18 at 4:49
Christopher MarleyChristopher Marley
1,020115
Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
– Kayla Martin
Nov 22 '18 at 4:54
@KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
– idea
Nov 22 '18 at 4:59
Okay, thank you!
– Kayla Martin
Nov 22 '18 at 5:01
add a comment |
Really, these polynomials target a single skill: taking out a common factor. For example, with $x^2+x$, both terms have $x$ as a factor, so you can divide it out (inverse distributive property of sorts) and get $x(x+1)$.
$$$$
$$x^3+x = x(x^2+1)$$
Roots are $0$, $i$, and $-i$.
$$x^4+x^2 = x^2(x^2+1)$$
Roots are $0$, $i$, and $-i$.
$$x^3-x = x(x^2-1) = x(x+1)(x-1)$$
Roots are $0$, $1$, and $-1$.
answered Nov 22 '18 at 4:49
Christopher MarleyChristopher Marley
1,020115
Really, these polynomials target a single skill: taking out a common factor. For example, with $x^2+x$, both terms have $x$ as a factor, so you can divide it out (inverse distributive property of sorts) and get $x(x+1)$.
$$$$
$$x^3+x = x(x^2+1)$$
Roots are $0$, $i$, and $-i$.
$$x^4+x^2 = x^2(x^2+1)$$
Roots are $0$, $i$, and $-i$.
$$x^3-x = x(x^2-1) = x(x+1)(x-1)$$
Roots are $0$, $1$, and $-1$.
answered Nov 22 '18 at 4:49
Christopher MarleyChristopher Marley
1,020115
answered Nov 22 '18 at 4:49
Christopher MarleyChristopher Marley
1,020115
answered Nov 22 '18 at 4:49
Christopher MarleyChristopher Marley
1,020115
answered Nov 22 '18 at 4:49
Christopher MarleyChristopher Marley
1,020115
1,020115
Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
– Kayla Martin
Nov 22 '18 at 4:54
@KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
– idea
Nov 22 '18 at 4:59
Okay, thank you!
– Kayla Martin
Nov 22 '18 at 5:01
add a comment |
Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
– Kayla Martin
Nov 22 '18 at 4:54
@KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
– idea
Nov 22 '18 at 4:59
Okay, thank you!
– Kayla Martin
Nov 22 '18 at 5:01
Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
– Kayla Martin
Nov 22 '18 at 4:54
Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
– Kayla Martin
Nov 22 '18 at 4:54
@KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
– idea
Nov 22 '18 at 4:59
@KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
– idea
Nov 22 '18 at 4:59
Okay, thank you!
– Kayla Martin
Nov 22 '18 at 5:01
Okay, thank you!
– Kayla Martin
Nov 22 '18 at 5:01
add a comment |
I think it's important you understand this idea for more difficult cases than you ask about, so my answer will touch on a more general method for solving these problems, and show you how it works for slightly more difficult cases.
We have that if for some polynomial $p(x)$, $p(a)=0$, then:
$$p(x)=(x-a)cdot q(x)$$
Where $q(x)$ is a polynomial one degree smaller than $p(x)$ (the degree of a polynomial just means the largest power of $x$ in it).
If you can find any point where a polynomial $p(x) $ has a zero, you can take out the factor in this way and find the smaller polynomial $q(x)$, which can either be solved or reduced again.
Let me demonstrate first on (b)
We have that $p(x)=x^4-x^2$
I like to test values $0,1,-1,2,-2$ etc... Here, it's plain that $p (0)=0$, so we take out $(x-0)$, or just $x$. We now have that:
$$x^4-x^2=xcdot q(x)$$
From which we get that $q(x)=x^3-x$. But $q(0)=0$, so we can take out another factor $x$, giving $$x^4-x^2=x^2cdot r (x)$$ It follows that $r(x)=x^2+1$, a quadratic we can solve by the Quadratic Formula.
This is a rather bland example, so I'll demonstrate the procedure on a better example.
Suppose we have $$f(x)=x^3+4x^2+7x+4$$
Plugging in values for $x$, we get that $f(-1)=0$
So we have that:
$$x^3+4x^2+7x+4=(x+1)cdot g(x)$$
Or rather,
$$x^3+4x^2+7x+4=(x+1)(x^2+px+q)$$ for some real $p,q$.
Expanding this out we get:
$$x^3+4x^2+7x+4=x^3+(p+1)x^2+(p+q)x+q$$
From which it follows that $p=3, q=4$.
So: $$f(x)=(x+1)(x^2+3x+4)$$, which is a quadratic we can solve (even if the solutions aren't real)
answered Nov 22 '18 at 5:50
Rhys HughesRhys Hughes
5,0541427
add a comment |
I think it's important you understand this idea for more difficult cases than you ask about, so my answer will touch on a more general method for solving these problems, and show you how it works for slightly more difficult cases.
We have that if for some polynomial $p(x)$, $p(a)=0$, then:
$$p(x)=(x-a)cdot q(x)$$
Where $q(x)$ is a polynomial one degree smaller than $p(x)$ (the degree of a polynomial just means the largest power of $x$ in it).
If you can find any point where a polynomial $p(x) $ has a zero, you can take out the factor in this way and find the smaller polynomial $q(x)$, which can either be solved or reduced again.
Let me demonstrate first on (b)
We have that $p(x)=x^4-x^2$
I like to test values $0,1,-1,2,-2$ etc... Here, it's plain that $p (0)=0$, so we take out $(x-0)$, or just $x$. We now have that:
$$x^4-x^2=xcdot q(x)$$
From which we get that $q(x)=x^3-x$. But $q(0)=0$, so we can take out another factor $x$, giving $$x^4-x^2=x^2cdot r (x)$$ It follows that $r(x)=x^2+1$, a quadratic we can solve by the Quadratic Formula.
This is a rather bland example, so I'll demonstrate the procedure on a better example.
Suppose we have $$f(x)=x^3+4x^2+7x+4$$
Plugging in values for $x$, we get that $f(-1)=0$
So we have that:
$$x^3+4x^2+7x+4=(x+1)cdot g(x)$$
Or rather,
$$x^3+4x^2+7x+4=(x+1)(x^2+px+q)$$ for some real $p,q$.
Expanding this out we get:
$$x^3+4x^2+7x+4=x^3+(p+1)x^2+(p+q)x+q$$
From which it follows that $p=3, q=4$.
So: $$f(x)=(x+1)(x^2+3x+4)$$, which is a quadratic we can solve (even if the solutions aren't real)
answered Nov 22 '18 at 5:50
Rhys HughesRhys Hughes
5,0541427
add a comment |
I think it's important you understand this idea for more difficult cases than you ask about, so my answer will touch on a more general method for solving these problems, and show you how it works for slightly more difficult cases.
We have that if for some polynomial $p(x)$, $p(a)=0$, then:
$$p(x)=(x-a)cdot q(x)$$
Where $q(x)$ is a polynomial one degree smaller than $p(x)$ (the degree of a polynomial just means the largest power of $x$ in it).
If you can find any point where a polynomial $p(x) $ has a zero, you can take out the factor in this way and find the smaller polynomial $q(x)$, which can either be solved or reduced again.
Let me demonstrate first on (b)
We have that $p(x)=x^4-x^2$
I like to test values $0,1,-1,2,-2$ etc... Here, it's plain that $p (0)=0$, so we take out $(x-0)$, or just $x$. We now have that:
$$x^4-x^2=xcdot q(x)$$
From which we get that $q(x)=x^3-x$. But $q(0)=0$, so we can take out another factor $x$, giving $$x^4-x^2=x^2cdot r (x)$$ It follows that $r(x)=x^2+1$, a quadratic we can solve by the Quadratic Formula.
This is a rather bland example, so I'll demonstrate the procedure on a better example.
Suppose we have $$f(x)=x^3+4x^2+7x+4$$
Plugging in values for $x$, we get that $f(-1)=0$
So we have that:
$$x^3+4x^2+7x+4=(x+1)cdot g(x)$$
Or rather,
$$x^3+4x^2+7x+4=(x+1)(x^2+px+q)$$ for some real $p,q$.
Expanding this out we get:
$$x^3+4x^2+7x+4=x^3+(p+1)x^2+(p+q)x+q$$
From which it follows that $p=3, q=4$.
So: $$f(x)=(x+1)(x^2+3x+4)$$, which is a quadratic we can solve (even if the solutions aren't real)
answered Nov 22 '18 at 5:50
Rhys HughesRhys Hughes
5,0541427
I think it's important you understand this idea for more difficult cases than you ask about, so my answer will touch on a more general method for solving these problems, and show you how it works for slightly more difficult cases.
We have that if for some polynomial $p(x)$, $p(a)=0$, then:
$$p(x)=(x-a)cdot q(x)$$
Where $q(x)$ is a polynomial one degree smaller than $p(x)$ (the degree of a polynomial just means the largest power of $x$ in it).
If you can find any point where a polynomial $p(x) $ has a zero, you can take out the factor in this way and find the smaller polynomial $q(x)$, which can either be solved or reduced again.
Let me demonstrate first on (b)
We have that $p(x)=x^4-x^2$
I like to test values $0,1,-1,2,-2$ etc... Here, it's plain that $p (0)=0$, so we take out $(x-0)$, or just $x$. We now have that:
$$x^4-x^2=xcdot q(x)$$
From which we get that $q(x)=x^3-x$. But $q(0)=0$, so we can take out another factor $x$, giving $$x^4-x^2=x^2cdot r (x)$$ It follows that $r(x)=x^2+1$, a quadratic we can solve by the Quadratic Formula.
This is a rather bland example, so I'll demonstrate the procedure on a better example.
Suppose we have $$f(x)=x^3+4x^2+7x+4$$
Plugging in values for $x$, we get that $f(-1)=0$
So we have that:
$$x^3+4x^2+7x+4=(x+1)cdot g(x)$$
Or rather,
$$x^3+4x^2+7x+4=(x+1)(x^2+px+q)$$ for some real $p,q$.
Expanding this out we get:
$$x^3+4x^2+7x+4=x^3+(p+1)x^2+(p+q)x+q$$
From which it follows that $p=3, q=4$.
So: $$f(x)=(x+1)(x^2+3x+4)$$, which is a quadratic we can solve (even if the solutions aren't real)
answered Nov 22 '18 at 5:50
Rhys HughesRhys Hughes
5,0541427
answered Nov 22 '18 at 5:50
Rhys HughesRhys Hughes
5,0541427
answered Nov 22 '18 at 5:50
Rhys HughesRhys Hughes
5,0541427
answered Nov 22 '18 at 5:50
Rhys HughesRhys Hughes
5,0541427
5,0541427
add a comment |
add a comment |
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What have you tried? Where are you confused?
– Dzoooks
Nov 22 '18 at 4:45