Math, asked by loganm2003, 6 months ago

Find a polynomial function of least degree having only real​ coefficients, a leading coefficient of​ 1, and zeros of and The polynomial function is ​f(x) nothing. ​(Simplify your​ answer.)

Answers

Answered by nibha028128
1

Answer:

The Rational Zero Theorem tells us that if \displaystyle \frac{p}{q}

q

p

is a zero of \displaystyle f\left(x\right)f(x), then p is a factor of –1 and q is a factor of 4.

{

p

q

=

factor of constant term

factor of leading coefficient

=

factor of -1

factor of 4

The factors of –1 are \displaystyle \pm 1±1 and the factors of 4 are \displaystyle \pm 1,\pm 2±1,±2, and \displaystyle \pm 4±4. The possible values for \displaystyle \frac{p}{q}

q

p

are \displaystyle \pm 1,\pm \frac{1}{2}±1,±

2

1

, and \displaystyle \pm \frac{1}{4}±

4

1

.

These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1.

Synthetic division with 1 as the divisor and {4, 0, -3, -1} as the quotient. Solution is {4, 4, 1, 0}

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