coefficient of x³ in -4x³+x²-9x³+4x+2
Answers
Step-by-step explanation:
Simple rule for factorisation is that if sum of coefficients is 0 then (x-1) is a factor. If sum of coefficients of even powers and odd powers is the same then (x+1) is a factor.
General rule is if polynomial evaluates to 0 for x=a then (x-a) is a factor. This is a well-known theorem.
For the polynomial, coefficients of even powers are 2, 14, 2 (add to 18) and coefficients of even powers is -9, -9 (add to -18).
Sum of all coefficients add to 0 hence, (x-1) is a factor.
2x⁴ – 9x³ + 14 x² - 9x + 2
=2x^4 –2x^3–7x^3 +7x^2+7x^2 –7x-2x +2
=2x^4 –2x^3–7x^3 +7x^2+7x^2 –7x-2x +2
=2x^3(x-1) -7x^2(x-1) +7x(x-1) -2(x-1)
=(x-1) (2x^3 -7x^2+7x-2)
For polynomial (2x^3 -7x^2+7x-2), sum of coefficients is 0; again (x-1) is a factor.
2x^3 -7x^2+7x-2
=2x^3 -2x^2–5x^2 +5x+2x -2
=2x^3-2x^2 –5x^2 +5x +2x -2
=2x^2(x-1) -5x(x-1)+2(x-1)
=(x-1)(2x^2–5x+2)
Now for 2x^2–5x+2, if we take x=2, polynomial evaluates to 2×(2^2)-5×2+2=0 hence (x-2) is a factor.
2x^2–5x+2
=2x^2–4x-x+2
=2x(x-2)-1(x-2)
=(x-2)(2x-1)
Putting everything together
2x⁴ – 9x³ + 14 x² - 9x + 2
=(x-1) (2x^3 -7x^2+7x-2)
=(x-1)(x-1)(2x^2–5x+2)
=((x-1)^2)(x-2)(2x-1)
Ans.:((x-1)^2)(x-2)(2x-1)
p.s.: This time, I have adhered to “if p(a)=0 then (x-a) is a factor (even for quadratic polynomial).”
first we solve the equation
-13x³+x²+4x+2
coefficient of x³ is
-13
answer is -13
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