Please explain how to factorise a quadratic or cubic polynomial using factor theorem.
Answers
◾️explain how to factorise a quadratic or cubic polynomial using factor theorem.
We know how to factorise quadratic equations already, either by observation or by use of the quadratic formula to find the solutions to f(x)=0 and therefore the factors. The quadratic formula allows us to solve and factorise any quadratic equation we may come across.
The factor theorem helps us to solve/factorise cubic functions.
There is a similar formula to the quadratic formula called the cubic formula which allows any cubic equation to be solved but it is incredibly long and time consuming.
So, instead of trying to learn the cubic formula, the factor theorem allows us to find the first solution and therefore first factor to a cubic, leaving us with a binomial and a quadratic, from which we can fully factorise.
The way the factor theorem works is by finding values of x that cause f(x) [in this case, the cubic] to equal zero.
This is done unelegantly by trial and error, however all the cubics we will come across will have simple solutions, so very few trials will be needed.
After we find a value of x that equates the cubic to zero, call it a, we know that a factor of the cubic must be (x-a). This gives us the first factor, leaving the cubic as a binomial and a quadratic, so we can easily fully factorise it Let's look at an example : Let's take the cubic equation f(x)=x3-6x2+11x-6.
We are going to test a few different values of x and see what values of f(x) they give us. We start with the easiest values : 0, 1, -1, 2, -2 etc. f(0) = 0-0+0-6 = -6, therefore (x-0) is not a factor. f(1) = 1-6+11-6 = 0, therefore (x-1) is a factor. We have found the first factor, so by dividing the cubic by (x-1) we get a quadratic which we can solve.
However, we can also keep going with the factor theorem. f(2) = 8-24+22-6 = 0 therefore (x-2) is also a factor. Knowing these two factors gives us the third factor easily. One way of seeing this is that -6 at the end of the cubic, already having a product of 2 from the other two factors, you must multiply by -3.
Long division is another way of spotting this.
Now we can use the factor theorem to factorise cubic equations.
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Answer:
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Question:
◾️explain how to factorise a quadratic or cubic polynomial using factor theorem.
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Answer:
We know how to factorise quadratic equations already, either by observation or by use of the quadratic formula to find the solutions to f(x)=0 and therefore the factors. The quadratic formula allows us to solve and factorise any quadratic equation we may come across.
The factor theorem helps us to solve/factorise cubic functions.
There is a similar formula to the quadratic formula called the cubic formula which allows any cubic equation to be solved but it is incredibly long and time consuming.
So, instead of trying to learn the cubic formula, the factor theorem allows us to find the first solution and therefore first factor to a cubic, leaving us with a binomial and a quadratic, from which we can fully factorise.
The way the factor theorem works is by finding values of x that cause f(x) [in this case, the cubic] to equal zero.
This is done unelegantly by trial and error, however all the cubics we will come across will have simple solutions, so very few trials will be needed.
After we find a value of x that equates the cubic to zero, call it a, we know that a factor of the cubic must be (x-a). This gives us the first factor, leaving the cubic as a binomial and a quadratic, so we can easily fully factorise it Let's look at an example : Let's take the cubic equation f(x)=x3-6x2+11x-6.
We are going to test a few different values of x and see what values of f(x) they give us. We start with the easiest values : 0, 1, -1, 2, -2 etc. f(0) = 0-0+0-6 = -6, therefore (x-0) is not a factor. f(1) = 1-6+11-6 = 0, therefore (x-1) is a factor. We have found the first factor, so by dividing the cubic by (x-1) we get a quadratic which we can solve.
However, we can also keep going with the factor theorem. f(2) = 8-24+22-6 = 0 therefore (x-2) is also a factor. Knowing these two factors gives us the third factor easily. One way of seeing this is that -6 at the end of the cubic, already having a product of 2 from the other two factors, you must multiply by -3.
Long division is another way of spotting this.
Now we can use the factor theorem to factorise cubic equations.
_________________________
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