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swat32
3 years ago
12

Write a polynomial function f(x) (in factored form AND standard form) of least degree that has rational coefficients, a leading

coefficient of 1, and -4, 1, and 2 as zeros.
Bit confused about the process. Any help would be appreciated!
Mathematics
2 answers:
yawa3891 [41]3 years ago
5 0

Answer:

f(x) = (x+4)(x-1)(x-2)

Step-by-step explanation:

We are given the roots as -4, 1, and 2

We are also given the leading coefficient as 1

We can write the polynomial as

f(x)=  a( x- r1)(x-r2) (x-r3) ...... where rn  are the roots  and a is the leading coefficient

f(x)= 1( x- -4) (x-1)(x-2)

    = 1(x+4) (x-1) (x-2)

We can drop the leading 1 since it is not needed

f(x) = (x+4)(x-1)(x-2)

zmey [24]3 years ago
4 0

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • To write a polynomial function f(x) x) (in factored form and standard form) of least degree that has rational coefficients, a leading coefficient of 1, and -4, 1, and 2 as zeros.

<u>Solution</u><u> </u><u>:</u><u>-</u>

Given zeros are ,

\longrightarrow Zeros = -4 , 1 and 2

Since there are three zeroes , the polynomial will be a <u>cubic</u><u> </u><u>polynomial</u><u> </u>.

We know that if , \alpha , \beta \ \& \ \gamma are the zeros of the cubic polynomial then , we can write the polynomial as ,

\sf\longrightarrow f(x) =k ( x -\alpha)(x-\beta)(x-\gamma)

where k is constant ,

Hence here ,

  • \alpha = -4
  • \beta = 1
  • \gamma = 2
  • k = 1

Hence here the polynomial can be written in <u>factored</u><u> </u><u>form</u><u> </u> as ,

\sf\longrightarrow f(x) =1 [ x -(-4)] ( x-1)(x-2)\\

\sf\longrightarrow \underset{\bf Factored\ form}{\underbrace{\underline{\underline{ f(x) = (x +4)(x-1)(x-2)}}}}

Again we know that the<u> </u><u>Standard</u><u> form</u><u> </u> of a cubic polynomial is ,

\sf\longrightarrow p(x) = ax^3+bx^2+ c x + d

Now to find in standard form multiply the all three , as ,

\sf\longrightarrow f(x) = (x+4)(x-1)(x-2)\\

\sf\longrightarrow f(x) = [ x ( x-1) +4(x-1)] (x-2)\\

\sf\longrightarrow f(x) = [ x^2-x +4x -4](x-2)\\

\sf\longrightarrow  f(x) = ( x^2+3x-4)(x-2)\\

\sf\longrightarrow  f(x) = x( x^2+3x-4)-2(x^2+3x-4) \\

\sf\longrightarrow f(x) = x^3+3x^2-4x -2x^2-6x+8\\

Add like terms ,

\sf\longrightarrow \underset{\bf Standard\ form}{\underbrace{\underline{\underline{f(x) = x^3+x^2-10x+8}}}}

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