Question

find a cubic polynomial p(X) whose zeroes are the same as those colLectively of polynomials g1(X)=2x2 -9x+4 and g2(X) = 2x2+3x-2.given that p(0)=8

Answer 1

SOLUTION

TO DETERMINE

The cubic polynomial p(x) whose zeroes are the same as those collectively of the below polynomials given that p(0) = 8

$\sf{g_1(x) = 2 {x}^{2} - 9x + 4 }$

$\sf{g_2(x) = 2 {x}^{2} + 3x - 2 }$

EVALUATION

Here the given polynomials are

$\sf{g_1(x) = 2 {x}^{2} - 9x + 4 }$

$\sf{g_2(x) = 2 {x}^{2} + 3x - 2 }$

We now find the zeroes of the polynomials

For first polynomial

$\sf{g_1(x) = 0 }$

$\sf{ \implies 2 {x}^{2} - 9x + 4 = 0}$

$\sf{ \implies 2 {x}^{2} - 8x - x + 4 = 0}$

$\sf{ \implies (x - 4)(2x - 1) = 0}$

$\displaystyle \sf{ \implies \: x = 4 \: ,\: \frac{1}{2} }$

So the zeroes of the first polynomial are

$\displaystyle \sf{ 4 \: ,\: \frac{1}{2} }$

Similarly the zeroes of the second polynomial are

$\displaystyle \sf{ - 2 \: ,\: \frac{1}{2} }$

So the zeroes of the required polynomial are

$\displaystyle \sf{ \ - 2 \:, \: 4 \: ,\: \frac{1}{2} }$

So the required cubic polynomial is of the form

$\displaystyle \sf{p(x) = a(x + 2)(x - \frac{1}{2})(x - 4) }$

Where a is the non zero constant to be determined

Now it is given that p(0) = 8

Which gives

$\displaystyle \sf{a(0 + 2)(0 - \frac{1}{2})(0 - 4) = 8 }$

$\displaystyle \sf{ \implies \: 4a = 8 }$

$\displaystyle \sf{ \implies \: a = 2 }$

Hence the required cubic polynomial is

$\displaystyle \sf{p(x) = 2(x + 2)(x - \frac{1}{2})(x - 4) }$

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Answer 2

Answer:

Step-by-step explanation: