Question

x^4- 6x³ + 7x² + 6x-8

Answer 1

p(x)=x⁴-6x³+7x²+6x-8

If x=1,

p(1)=1⁴-6(1)³+7(x)²+6(1)-8

=1-6+7+6-8

p(1)=0.

so, x-1 is a factor of p(x).

Now, By dividing p(x) by x-1:

x-1)x⁴ -6x³+7x²+6x-8(x³-5x²+2x+8

(-)x⁴(+)-x³

-5x³ +7x²

(+)-5x³(-)5x²

2x² +6x

(-)2x²(+)-2x

8x -8

(-)8x(+)-8

0.

q(x)=x³-5x²+2x+8

If, x=1,

p(x)=(1)³-5(1)²+2(1)+8

=1-5+2+8=6>0

If x=-1,

P(-1)=(-1)³-5(-1)²+2(-1)+8

=-1-5-2+8=-8+8=0

so, x+1 is a factor of p(x)=q(x)=x³-5x²+2x+8.

x+1)x³ -5x²+2x+8(x²-6x+8

(-)x³(-)1x²

-6x² 2x

(+)-6x²(+)-6x

8x+8

(-)8x(-)8

0.

q(x)=x²-6+8=0

x²-2x-4x+8=0

x(x-2)-4(x-2)=0

(x-2)(x-4)=0

x=2,4.

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