Line JK passes through points J(–4, –5) and K(–6, 3). If the equation of the line is written in slope-intercept form, y = mx + b
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Answer:
x^4 -53x^2 +108x +160
Step-by-step explanation:
If <em>a</em> is a zero, then (<em>x-a</em>) is a factor. For the given zeros, the factors are ...
p(x) = (x +8)(x +1)(x -4)(x -5)
Multiplying these out gives the polynomial in standard form.
= (x^2 +9x +8)(x^2 -9x +20)
We note that these factors have a sum and difference with the same pair of values, x^2 and 9x. We can use the special form for the product of these to simplify our working out.
= (x^2 +9x)(x^2 -9x) +20(x^2 +9x) +8(x^2 -9x) +8(20)
= x^4 -81x^2 +20x^2 +180x +8x^2 -72x +160
p(x) = x^4 -53x^2 +108x +160
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The graph shows this polynomial has the required zeros.
