Given that y=3x3+7x2−48x+49 and that y has the same remainder when it is divided by x+k or x−k, find the possible values of k.

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clarkchica44klt · Accepted Answer

Step 1f(−k)=f(k)3(−k)3+7(−k)2−48(−k)+49=3(k)3+7(k)2−48(k)+49bf−3k3−7k2+48k+49=3k3+7k2−48k+49−3k3−−3k3−7k2−7k2+48k+48k+49−49=0−6k3+14k2+96k=0−6k3k+14k2k+96kk=0k6k2+14k−96=0Right there. 7(−k)2=7k2.Looking through it, there are more sign errors throughout.−7k2−7k2would be −14k2, not 14k2Another one: −3k3−−3k3=0Dividing by k isn&#039;t exactly a kosher operation. What if k=0

Chad Robinson · Accepted Answer

Step 1By the factor theorem,3x3+7x2−48x+49=(ax+b)(x2−k2)+cIn other words, dividing by a quadratic term leaves a linear quotient and a constant remainder, for the powers on both sides to match.Observe that there is only one way to make the terms of x3,x2,xThus ax2=3x3⇒x=3 and hence−ak2=−48⇒k2=16, k=±4

Given that \(\displaystyle{y}={3}{x}^{{3}}+{7}{x}^{{2}}-{48}{x}+{49}\) and that y has

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