Question

hw 1 - algebra for calculus section 0.1: problem 1 (1 point)
(a) factor the expression 4x² + 25x - 21.
help (formulas)
(b) solve the equation 4x² + 25x - 21 = 0. if there is more than one correct answer, enter your answers as a comma separated list. if there are no solutions, enter none.
x = help (numbers)
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Explanation:

Step1: Factor the quadratic expression

For a quadratic expression \(ax^{2}+bx + c\) (here \(a = 4\), \(b=25\), \(c=-21\)), we need to find two numbers \(m\) and \(n\) such that \(m\times n=ac=- 84\) and \(m + n=b = 25\). The numbers are \(28\) and \(-3\) since \(28\times(-3)=-84\) and \(28+( - 3)=25\). Then we rewrite the middle - term: \(4x^{2}+25x - 21=4x^{2}+28x-3x - 21\). Group the terms: \((4x^{2}+28x)-(3x + 21)=4x(x + 7)-3(x + 7)=(4x - 3)(x+7)\).

Step2: Solve the quadratic equation

Set \((4x - 3)(x + 7)=0\). According to the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we have \(4x-3=0\) or \(x + 7=0\). For \(4x-3=0\), we solve for \(x\) and get \(x=\frac{3}{4}\). For \(x + 7=0\), we solve for \(x\) and get \(x=-7\).

Answer:

(a) \((4x - 3)(x + 7)\)
(b) \(\frac{3}{4},-7\)