Question

Check whether (x²-11x+28) is a factor of the polynomial x³-12x²+39x-28


please answer step wise please guys

$1$
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Answer 1

Step-by-step explanation:

1 is the answer ok na dude

Answer 2

Answer:

Solving (x²-11x+28):-

x²-11x+28

Product=28

Sum= -11

Numbers= -7,-4

=x²-7x-4x+28)

= x(x-7)-4(x-7)

= (x-4)(x-7)

x-4=0 or x-7=0

x=4,7

Now (x²-11x+28) will be factor of:-

${x}^{3} - 12 {x}^{2} + 39x - 28$

only if x-4=0 and x-7=0 are factors of the given polynomial then only (x²-11x+28) will be it's factor.

Checking x=4

${4}^{3} - 12 { \times 4}^{2} + 39(2) - 28 = 0$

$64 - 12 \times 16 + 78 - 28 = 0$

$64 - 192 + 78 - 28 = 0$

= 142-220=0

= -78 not equal to 0

Checking x=7

${7}^{3} - 12 \times {7}^{2} + 39 \times 7 - 28 = 0$

343-12×49+273-28=0

= 343-588+273-28=0

= 28-28=0

0=0

Since, x-4 is not a factor of the given polynomial hence, (x²-11x+28) is not a factor.

For verification I also attached the solution.