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Answer:
3 + 11a³ - 7a² + 18a - 18
Step-by-step explanation:
<u>When multiplying with two brackets, you need to multiply the three terms, (a²), (4a) and (-6) from the first bracket to all the terms in the second brackets, (3a²), (-a) and (3) individually. I have put each multiplied term in a bracket so it is easier.</u>
(a² + 4a - 6) × (3a² - a + 3) =
(a² × <em>3a²</em>) + {a² × <em>(-a)</em>} + (a² × <em>3</em>) + (4a × <em>3a²</em>) + {4a × <em>(-a)</em>} + (4a × <em>3</em>) + {(-6) × <em>a²</em>) + {(-6) × <em>(-a)</em>} + {(-6) × <em>3</em>}
<u>Now we can evaluate the terms in the brackets. </u>
(a² × 3a²) + {a² × (-a)} + (a² × 3) + (4a × 3a²) + {4a × (-a)} + (4a × 3) + {(-6) × a²) + {(-6) × (-a)} + {(-6) × 3} =
3 + (-a³) + 3a² + 12a³ + (-4a²) + 12a + (-6a²) + 6a + (-18)
<u>We can open the brackets now. One plus and one minus makes a minus. </u>
3 + (-a³) + 3a² + 12a³ + (-4a²) + 12a + (-6a²) + 6a + (-18) =
3 -a³ + 3a² + 12a³ -4a² + 12a -6a² + 6a -18
<u>Evaluate like terms.</u>
3 -a³ + 3a² + 12a³ -4a² + 12a -6a² + 6a -18 = 3
+ 11a³ - 7a² + 18a - 18
To solve for p, square both sides and move the equation to one side is equal to zero. Then factor. I chose -2 and -5 because -2 x -5 = 10 and -2 + -5 = -7. After factoring, solve p-5=0 and p-2=0. When one of those equals zero then the equation (p-5)(p-2)=0 is true.
