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ex #1: h(x) = (9x^5 + 4)(7x - 5) =

Question

ex #1: h(x) = (9x^5 + 4)(7x - 5) =

Explanation:

Step1: Apply Distributive Property (FOIL)

Multiply each term in the first binomial by each term in the second binomial:
$(9x^5 + 4)(7x - 5) = 9x^5 \cdot 7x + 9x^5 \cdot (-5) + 4 \cdot 7x + 4 \cdot (-5)$

Step2: Simplify Each Term

  • For $9x^5 \cdot 7x$: Multiply coefficients and add exponents of $x$: $9 \cdot 7 = 63$, $x^5 \cdot x = x^{5+1} = x^6$, so $63x^6$.
  • For $9x^5 \cdot (-5)$: Multiply coefficients: $9 \cdot (-5) = -45$, so $-45x^5$.
  • For $4 \cdot 7x$: Multiply coefficients: $4 \cdot 7 = 28$, so $28x$.
  • For $4 \cdot (-5)$: Multiply: $4 \cdot (-5) = -20$.

Step3: Combine Terms

Combine the simplified terms: $63x^6 - 45x^5 + 28x - 20$

Answer:

$63x^6 - 45x^5 + 28x - 20$