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klasskru [66]
2 years ago
9

Describe the first step when factoring any polynomial. Why is this the first step? Explain.

Mathematics
2 answers:
shtirl [24]2 years ago
4 0

Answer with explanation:

→Consider a Polynomial

   \rightarrow ax^n+bx^{n-1}+cx^{n-2}+...........+d

→We will use rational root theorem to find out the factors of the polynomial.

  \rightarrow ax^n+bx^{n-1}+cx^{n-2}+...........+d\\\\\rightarrow a(x^n+\frac{bx^{n-1}}{a} +\frac{cx^{n-1}}{a}+..........+\frac{d}{a})

→Factors of d are

        =\pm 1 , .....\pm d\\\\ \text{Factors of a are}=\pm 1,.....\pm a\\\\ \pm\frac{d}{a}  

⇒Now,substitute these integers that is factors into Polynomial expression to find which expression is equal to Zero.These integers will be factor of Polynomial expression.

For example

 f(x)=x²-3 x+2

By rational root theorem ,roots of the polynomial can be , -1,1,2,-2.

f(1)=1²-3×1+2

   =1-3+2

   =0

f(2)=2²-3×2+2

   =4-6+2

  =0

So, 1 and 2 are root of the polynomial expression.

seropon [69]2 years ago
3 0
Step 1: If there is a common factor, factor out the GCF. Step 2<span>: Identify the number of terms: (i) If polynomial has two terms, convert polynomial into difference of two squares or sum of two cubes or difference of two cubes.</span>
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Answer:

Step-by-step explanation:

Hello, please consider the following.

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Hope this helps.

Do not hesitate if you need further explanation.

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a good example for that will be √(4), well, (2)(2) is 4, so 2 is a root, but (-2)(-2) is also 4, therefore -2 is also a root, so you'd always get a pair of valid roots from an even root, like 2 or 4 or 6 and so on.

therefore, complex solutions or roots are never by their lonesome, their sister the conjugate is always with them, so if there's a root a + bi, her sister a - bi is also coming along too.

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