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ivolga24 [154]
1 year ago
11

Write a polynomial function, p(x), with degree 3 that has p(7) =0

Mathematics
1 answer:
VikaD [51]1 year ago
3 0

The polynomial function, p(x), with degree 3 that has p(7) = 0 is p(x) = x^{3} -7x^{2} +12x-84=0.

According to the question,

We have the following conditions to find the polynomial function, p(x):

The degree has to be 3 and the value of p(7) should be 0.

Now, we are sure that we have one term as x^{3}.

Now, when 7 has to be multiplied three times we have 343 as the result.

So, we will try to make it zero in the next term.

The next term can be-7x^{2} because we will get -343 and the result of the first two terms will be 0.

Now, the third term can be 12x (you can take any term but we have to make sure that the end result is 0).

Now, the result will be 84 when we put 7 in place of x.

Now, we can have -84.

So, we will add these 4 terms to form the polynomial function:

p(x) = x^{3} -7x^{2} +12x-84=0

Hence, the required polynomial function is p(x) = x^{3} -7x^{2} +12x-84=0.

To know more about polynomial function here

brainly.com/question/12976257

#SPJ1

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All the answers are given below.

Step-by-step explanation:

(i) (x+3)(x-3) =x² -3² =x² -9 {Since (a+b)(a-b) =a² -b²}

(ii) (x²-3)(x²+3) =(x²)² -3² =x^{4} -9 {Since (a+b)(a-b) =a² -b²}

(iii) (x^{15} +3)(x^{15}-3) =(x^{15} )^{2}-3^{2}  =x^{30}  -9

(iv) (x-3)(x²-9)(x+3) =(x²-9)(x²-9)=(x²-9)²=x^{4}-18x^{2} +81{Since, (a-b)² =a²-2ab+b²}

(v) (x²+y²)(x²-y²) = (x²)² -(y²)² =x^{4}-y^{4}

(vi) (x²+y²)² =x^{4}+2x^{2}  y^{2} +y^{4} {Since (a+b)² =a²+2ab+b²}

(vii) (x-y)²(x+y)² =(x²-y²)² =x^{4} -2x^{2} y^{2}+y^{4} {Since (a-b)² =a²-2ab+b² and (a+b)(a-b) =a²-b²}

(viii) (x-y)²(x²+y²)²(x+y)² =(x²+y²)²(x²-y²)² =(x^{4} -y^{4} )^{2} =x^{8}-2x^{4}  y^{4} +y^{8} (Answer)

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