9514 1404 393
Answer:
C. 15x^4 -29x^3 +6x^2 +8x
Explanation:
The answer choices differ in their coefficients of x^3 and x^2, so that is where we can focus our attention.
The multiplier only has x^2 and x terms in it, so the partial products that result in x^3 will be the x and x^2 terms of the multiplicand, respectively. That is, the x^3 term will be ...
(5x^2)(-7x) +(2x)(3x^2) = x^3(-35 +6) = -29x^3 . . . . . matches choices A and C
Similarly, the x^2 term will be ...
(5x^2)(4) +(2x)(-7x) = x^2(20 -14) = 6x^2 . . . . . matches only choice C
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Additional comment
In general, multiplying polynomials is an exercise in the use of the distributive property. Basically, every term of one polynomial is multiplied by every term of the other, and the like terms collected from the list of partial products.
When the product is of two quadratics, there are several partial products that get added to make any term except the first and last. Consider ...
(ax^2 +bx +c)(dx^2 +ex +g)
The x^4 term will be adx^4
The x^3 term will be (ae +bd)x^3
The x^2 term will be (ag +be +cd)x^2
The x term will be (bg +ce)x
The constant term will be cg.
For the product in this problem, g=0, and we're only concerned with the x^3 and x^2 terms in order to make a correct answer choice.