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Chapter 0: Problem 27

Find each product. $$\left(5 x^{2}-4\right)\left(3 x^{2}-7\right)$$

Short Answer

Expert verified
The product of the two binomials \((5 x^{2}-4)\) and \((3 x^{2}-7)\) is \(15x^4 -47x^2 +28\).

Step by step solution

01

Multiply the first terms

Multiply the first terms in each binomial: \(5x^2\) and \(3x^2\). This produces \(15x^4\).
02

Multiply the outside terms

Multiply the outside terms: \(5x^2\) from the first binomial and \(-7\) from the second. This produces \(-35x^2\).
03

Multiply the inside terms

Multiply the inside terms: \(-4\) from the first binomial and \(3x^2\) from the second. This gives \(-12x^2\).
04

Multiply the last terms

Multiply the last terms in each binomial: \(-4\) and \(-7\). This gives \(28\).
05

Add up all the products

Add up all the products obtained from the above steps to get the result. This results in \(15x^4 -35x^2 -12x^2 +28 = 15x^4 -47x^2 +28\).

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