Examkin

Soru

Zorluk: OrtaOperations on Polynomials

The position of a particle moving along a straight line is described by the polynomial s(t)=(2t2−3t+4)(t+3)s(t) = (2t^2 - 3t + 4)(t + 3), where tt represents the time in seconds and s(t)s(t) represents the displacement in meters. When the polynomial is expanded and written in the standard form at3+bt2+ct+dat^3 + bt^2 + ct + d, what is the value of b+cb + c?

Cevap: -2

Cevap

The value of b+cb + c is −2-2.
Expanding the polynomial s(t)=(2t2−3t+4)(t+3)s(t) = (2t^2 - 3t + 4)(t + 3) yields 2t3+3t2−5t+122t^3 + 3t^2 - 5t + 12. Comparing this to the standard form at3+bt2+ct+dat^3 + bt^2 + ct + d, we find b=3b = 3 and c=−5c = -5. Summing these coefficients gives b+c=3+(−5)=−2b + c = 3 + (-5) = -2.

Adım Adım Çözüm

1
Multiply each term of the trinomial 2t2−3t+42t^2 - 3t + 4 by each term of the binomial t+3t + 3.
2t3+6t2−3t2−9t+4t+122t^3 + 6t^2 - 3t^2 - 9t + 4t + 12
Apply the distributive property to expand the product of the polynomials.
2
Combine the like terms (the t2t^2 terms and the tt terms).
2t3+3t2−5t+122t^3 + 3t^2 - 5t + 12
Group and add the coefficients of like terms to simplify the polynomial.
3
Identify the coefficients bb (for the t2t^2 term) and cc (for the tt term) and find their sum.
b=3b = 3 and c=−5c = -5, so b+c=3+(−5)=−2b + c = 3 + (-5) = -2
Compare the simplified polynomial to the standard form at3+bt2+ct+dat^3 + bt^2 + ct + d to find the coefficients, then calculate the requested sum.

Anahtar Kavram

Operations on Polynomials

Alternatif Yöntem

Instead of expanding the whole expression, you can find the individual coefficients directly. The t2t^2 term is formed by (2t2)(3)+(−3t)(t)=6t2−3t2=3t2(2t^2)(3) + (-3t)(t) = 6t^2 - 3t^2 = 3t^2, so b=3b = 3. The tt term is formed by (−3t)(3)+(4)(t)=−9t+4t=−5t(-3t)(3) + (4)(t) = -9t + 4t = -5t, so c=−5c = -5. Thus, b+c=3−5=−2b + c = 3 - 5 = -2.
Tahmini Süre:1m 30s
Bu soruyu puanla

Konular

Operations on Polynomials

Bu konudan daha fazla soru