Question

f(x)=2x^4+x^3-14x^2-19x-6 if two of its zeros are-2 and -1

Answer 1

↑ Here is your answer ↓
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$(x+2)(x+1)$

$x(x+1) + 2(x+1)$

$x^2 + x + 2x + 2$

$x^2 + 3x + 2$

Now,

$2x^4+x^3-14x^2-19x-6 / x^2 + 3x + 2$

The quotient comes out as:

$2x^2 - 5x - 3 = 0$

By solving the qudratic equation,

$x = 3, x = -1/2$

Answer 2

Hey !!

Division is in the image..

Here is your answer...

$f(x) = {2x}^{4} + {x}^{3} - {14x}^{2} - 19x - 6 \\ \\ x = ( - 2) \: \: and \: \: x = ( - 1) \\ \\ x + 2 = 0 \: \: \: and \: \: \: x + 1 = 0 \\ \\ = (x + 2)(x + 1) \\ \\ = {x}^{2} + 2x + x + 2 \\ \\ = {x}^{2} + 3x + 2 \\ \\ dividing \: \: f(x) \: \: by \: \: {x}^{2} + 3x + 2 \: \: you \: \: will \: \: get \: \: ...... \\ \\ 2 {x}^{2} - 5x - 3 \\ \\ now \: \: factorize \: \: it \\ \\ \\ 2 {x}^{2} - 5x - 3 = 0 \\ \\ 2 {x}^{2} - 6x + x - 3 = 0 \\ \\ 2x(x - 3) + 1(x - 3) = 0 \\ \\ (2x + 1)(x - 3) = 0 \\ \\ (2x + 1) = 0 \: \: \: \: and \: \: \: (x - 3) = 0 \\ \\ x = \frac{ - 1}{2} \: \: \: and \: \: \: x = 3$


Hope it helps you...

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