How to solve this polynomial equation?

Question

How to solve this polynomial equation?

When S(x)= $x^{4} + a x^{3} + b x^{2} + 4 x - 7$ $x^4+ax^3+bx^2+4x-7$ is divided by $(x + 1)$ $(x+1)$ $(x - 1)$ $(x-1)$ , the remaider is $- 2 x + 5$ $-2x+5$ . Find a and b.

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Answer 1 · 2018-07-02T01:25:38

I get $a = - 6, b = - 13$ $a=-6, b=-13$ .

Explanation:

Since we know the remainder is $- 2 x + 5$ $-2x+5$ , we are sure that if we subtract that from the given polynomial then the difference will be exactly a multiple of $(x + 1) (x - 1) = x^{2} - 1$ $(x+1)(x-1)=x^2-1$ . So we render this:

$(x^{4} + a x^{3} + b x^{2} + 4 x - 7) - (- 2 x + 5) = x^{4} + a x^{3} + b x^{2} + 6 x - 12 = (x^{2} - 1) (x^{2} + p x + q)$ $(x^4+ax^3+bx^2+4x-7)-(-2x+5)=x^4+ax^3+bx^2+6x-12=(x^2-1)(x^2+px+q)$

Then we multiply out that last product and match like power terms:

$(x^{2} - 1) (x^{2} + p x + q) = x^{4} + p x^{3} + (q - 1) x^{2} - p x - q = x^{4} + a x^{3} + b x^{2} + 6 x - 12$ $(x^2-1)(x^2+px+q)=x^4+px^3+(q-1)x^2-px-q=x^4+ax^3+bx^2+6x-12$

Then match like powers:

$x^{4} = x^{4}$ $x^4=x^4$

$p x^{3} = a x^{3}$ $px^3=ax^3$

$b x^{2} = (q - 1) x^{2}$ $bx^2=(q-1)x^2$

$6 x = - p x$ $6x=-px$

$- 12 = q$ $-12=q$

So $p = - 6$ $p=-6$ meaning $a = p = - 6$ $a=p=-6$ , and $q = - 12$ $q=-12$ meaning $b = q - 1 = - 13$ $b=q-1=-13$ .

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How to solve this polynomial equation?

When S(x)= $x^{4} + a x^{3} + b x^{2} + 4 x - 7$ $x^4+ax^3+bx^2+4x-7$ is divided by $(x + 1)$ $(x+1)$ $(x - 1)$ $(x-1)$ , the remaider is $- 2 x + 5$ $-2x+5$ . Find a and b.

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How to solve this polynomial equation?

When S(x)= x4+ax3+bx2+4x−7x^4+ax^3+bx^2+4x-7 is divided by (x+1)(x+1)(x−1)(x-1), the remaider is −2x+5-2x+5. Find a and b.

1 Answer

Explanation:

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Impact of this question

When S(x)= $x^{4} + a x^{3} + b x^{2} + 4 x - 7$ $x^4+ax^3+bx^2+4x-7$ is divided by $(x + 1)$ $(x+1)$ $(x - 1)$ $(x-1)$ , the remaider is $- 2 x + 5$ $-2x+5$ . Find a and b.